Question 1.
If a young man drives his scooter at a speed of 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives the scooter at a speed of 40 km/hr, it produces air pollution and increases his expenditure on petrol to Rs 5 per km. He has a maximum of Rs 100 to spend on petrol and travel a maximum distance in one hour time with less pollution. Express this problem as an LPP and solve it graphically. What value do you find here?          (CBSE 2013)

\displaystyle \text{Answer:}
\displaystyle \text{Let the young man drive } x \text{ km at a speed of } 25 \text{ km/hr and } y \text{ km at a speed of } 40 \text{ km/hr.}
\displaystyle \text{Clearly,}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{It is given that, he spends Rs } 2 \text{ per km if he drives at a speed of } 25 \text{ km/hr and Rs } \\ 5 \text{ per km if he drives at a speed of } 40 \text{ km/hr.}
\displaystyle \text{Therefore, money spent by him when he travelled } x \text{ km and } y \text{ km are Rs } 2x \text{ and Rs } 5y \text{ respectively.}
\displaystyle \text{It is given that he has a maximum of Rs } 100 \text{ to spend.}
\displaystyle \text{Thus,}
\displaystyle 2x + 5y \le 100
\displaystyle \text{Time spent by him when travelling with a speed of } 25 \text{ km/hr } = \frac{x}{25} \text{ hr}
\displaystyle \text{Time spent by him when travelling with a speed of } 40 \text{ km/hr } = \frac{y}{40} \text{ hr}
\displaystyle \text{Also, the available time is } 1 \text{ hour.}
\displaystyle \frac{x}{25} + \frac{y}{40} \le 1
\displaystyle \text{Or,}
\displaystyle 40x + 25y \le 1000
\displaystyle \text{The distance covered is } Z = x + y \text{ which is to be maximized.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Maximize } Z = x + y
\displaystyle \text{Subject to:}
\displaystyle 2x + 5y \le 100
\displaystyle 40x + 25y \le 1000
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,20),\ \left(\frac{50}{3},\frac{40}{3}\right),\ (25,0).
\displaystyle \text{The values of } Z = x + y \text{ at these corner points are as follows:}
\displaystyle (0,0) \rightarrow 0
\displaystyle (0,20) \rightarrow 20
\displaystyle \left(\frac{50}{3},\frac{40}{3}\right) \rightarrow 30
\displaystyle (25,0) \rightarrow 25
\displaystyle \text{The maximum value of } Z \text{ is } 30 \text{ which is attained at } E.
\displaystyle \text{Thus, the maximum distance travelled by the young man is } 30 \text{ km, if he drives } \\ \frac{50}{3} \text{ km at a speed of } 25 \text{ km/hr and } \frac{40}{3} \text{ km at a speed of } 40 \text{ km/hr.}

Question 2.
A manufacturer has three machines installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must operate at least for 5 hours per day. He produces only two items, each requiring the use of three machines. The hours required for producing one unit each of the items on the three machines is given in the following table:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Item} & \text{Machine I} & \text{Machine II} & \text{Machine III} \\ \hline A & 1 & 2 & 1 \\ \hline B & 2 & 1 & \frac{5}{4} \\ \hline \end{array}

He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produce so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of item A and } y \text{ units of item B be manufactured. Therefore, } x,y \ge 0.
\displaystyle \text{As we are given,}
\displaystyle \text{Item}\quad \text{Number of hours required by the machine}
\displaystyle \text{I}\quad \text{II}\quad \text{III}
\displaystyle A:\ 1,\ 2,\ 1
\displaystyle B:\ 2,\ 1,\ \frac{5}{4}
\displaystyle \text{Machines I and II are capable of being operated for at most } 12 \text{ hours whereas} \\ \text{Machine III must operate at least for } 5 \text{ hours a day.}
\displaystyle \text{According to the question, the constraints are}
\displaystyle x + 2y \le 12
\displaystyle 2x + y \le 12
\displaystyle x + \frac{5}{4}y \ge 5
\displaystyle \text{He makes a profit of Rs } 6.00 \text{ on item A and Rs } 4.00 \text{ on item B.} \\ \text{Profit made by him in producing } x \text{ items of A and } y \text{ items of B is } 6x + 4y.
\displaystyle \text{Total profit } Z = 6x + 4y \text{ which is to be maximized.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Maximize } Z = 6x + 4y
\displaystyle \text{Subject to:}
\displaystyle x + 2y \le 12
\displaystyle 2x + y \le 12
\displaystyle x + \frac{5}{4}y \ge 5
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{First, we will convert the} \\ \text{inequalities into equations as follows:}
\displaystyle x + 2y = 12
\displaystyle 2x + y = 12
\displaystyle x + \frac{5}{4}y = 5
\displaystyle x = 0
\displaystyle y = 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,6),\ (4,4),\ (6,0),\ (5,0),\ (0,4).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle (0,6) \rightarrow 24
\displaystyle (4,4) \rightarrow 40
\displaystyle (6,0)\rightarrow 36
\displaystyle (5,0) \rightarrow 30
\displaystyle (0,4)\rightarrow 16
\displaystyle \text{The maximum value of } Z \text{ is } 40 \text{ which is attained at } (4,4).

Question 3.
Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants per day while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at minimum labour cost?

\displaystyle \text{Answer:}
\displaystyle \text{Let tailor A work for } x \text{ days and tailor B work for } y \text{ days.}
\displaystyle \text{In one day, A can stitch } 6 \text{ shirts and } 4 \text{ pants whereas B can stitch } 10 \text{ shirts and } 4 \text{ pants.}
\displaystyle \text{Thus in } x \text{ days, A can stitch } 6x \text{ shirts and } 4x \text{ pants whereas in } y \text{ days B can stitch } \\  10y \text{ shirts and } 4y \text{ pants.}
\displaystyle \text{It is given that the minimum requirement of the shirts and pants are respectively } 60 \text{ and } 32.
\displaystyle \text{Thus,}
\displaystyle 6x + 10y \ge 60
\displaystyle 4x + 4y \ge 32
\displaystyle \text{Further it is given that A and B earn Rs } 15 \text{ and Rs } \\ 20 \text{ per day respectively. Thus, A earns Rs }\\ 15x \text{ and B earns Rs } 20y.
\displaystyle \text{Let } Z \text{ denote the total cost.}
\displaystyle Z = 15x + 20y
\displaystyle \text{Days cannot be negative.}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{Minimize } Z = 15x + 20y
\displaystyle \text{Subject to:}
\displaystyle 6x + 10y \ge 60
\displaystyle 4x + 4y \ge 32
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{First we will convert inequalities} \\ \text{into equations as follows:}
\displaystyle 6x + 10y = 60
\displaystyle 4x + 4y = 32
\displaystyle x = 0
\displaystyle y = 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,8),\ (5,3),\ (10,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle (0,8) \rightarrow 160
\displaystyle (5,3) \rightarrow 135
\displaystyle A(10,0) \rightarrow 150
\displaystyle \text{The minimum value of } Z \text{ is } 135 \text{ which is attained at } (5,3).
\displaystyle \text{Thus, for minimum labour cost, A should work for } 5 \text{ days and B should work for } 3 \text{ days.}

Question 4.
A factory manufactures two types of screws, A and B, each type requiring the use of two machines — an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws A, while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws B. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 70 and screws B at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit. 

\displaystyle \text{Answer:}
\displaystyle \text{Let the factory manufacture } x \text{ screws of type A and } y \text{ screws of type B on each day.}
\displaystyle \text{Therefore, } x \ge 0 \text{ and } y \ge 0.
\displaystyle \text{The given information can be compiled in a table as follows:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Machine} & \text{Score A} & \text{Score B} & \text{Score C} \\ \hline \text{Automatic Machine (min)} & 4 & 6 & 4 \times 60 = 240 \\ \hline \text{Hand Operated Machine (min)} & 6 & 3 & 4 \times 60 = 240 \\ \hline \end{array}
\displaystyle 4x + 6y \le 240
\displaystyle 6x + 3y \le 240
\displaystyle \text{The manufacturer can sell a package of screws A at a profit of Rs } 0.7 \text{ and screws B at a profit of Re } 1.
\displaystyle \text{Total profit } Z = 0.7x + 1y
\displaystyle \text{The mathematical formulation of the given problem is}
\displaystyle \text{Maximize } Z = 0.7x + 1y
\displaystyle \text{Subject to:}
\displaystyle 4x + 6y \le 240
\displaystyle 6x + 3y \le 240
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{First we will convert the} \\ \text{inequalities into equations as follows:}
\displaystyle 4x + 6y = 240
\displaystyle 6x + 3y = 240
\displaystyle x = 0
\displaystyle y = 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (40,0),\ (30,20),\ (0,40).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z = 7x + 10y \\ \hline (40,0) & 280 \\ \hline (30,20) & 410 \\ \hline (0,40) & 400 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 410 \text{ at } (30,20).
\displaystyle \text{Thus, the factory should produce } 30 \text{ packages of screws A and } \\ 20 \text{ packages of screws B to get the maximum profit of Rs } 410.

Question 5.
A company produces two types of leather belts, say type A and B. Belt A is a superior belt, especially. Belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per type respectively. Each belt of type A requires twice as much leather as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.

How should the company manufacture the two types of belts in order to have a maximum overall profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let the company produce } x \text{ belts of type A and } y \text{ belts of type B.} \\ \text{Number of belts cannot be negative. Therefore, } x,y \ge 0.
\displaystyle \text{It is given that leather is sufficient only for } 800 \text{ belts per day (both A and B combined).}
\displaystyle \text{Therefore,}
\displaystyle x + y \le 800
\displaystyle \text{It is given that the rate of production of belts of type B is } 1000 \text{ per day.} \\ \text{Hence the time taken to produce } y \text{ belts of type B is } \frac{y}{1000}.
\displaystyle \text{And, since each belt of type A requires twice as much time as a belt of type B, the} \\ \text{rate of production of belts of type A is } 500 \text{ per day and therefore, total time} \\ \text{taken to produce } x \text{ belts of type A is } \frac{x}{500}.
\displaystyle \text{Thus, we have,}
\displaystyle \frac{x}{500} + \frac{y}{1000} \le 1
\displaystyle \text{Or,}
\displaystyle 2x + y \le 1000
\displaystyle \text{Belt A requires fancy buckle and only } 400 \text{ fancy buckles are available for this per day.}
\displaystyle x \le 400
\displaystyle \text{For belt of type B only } 700 \text{ buckles are available per day.}
\displaystyle y \le 700
\displaystyle \text{Profits on each type of belt are Rs } 2 \text{ and Rs } 1.50 \text{ per belt, respectively.} \\ \text{Therefore, profit gained on } x \text{ belts of type A and } y \text{ belts of type B is Rs }\\  2x \text{ and Rs } 1.50y \text{ respectively.}
\displaystyle \text{Hence, the total profit would be Rs } (2x + 1.50y). \text{ Let } Z \text{ denote the total profit.}
\displaystyle Z = 2x + 1.50y
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is;}
\displaystyle \text{Maximize } Z = 2x + 1.50y \text{Subject to:}
\displaystyle x + y \le 800
\displaystyle 2x + y \le 1000
\displaystyle x \le 400
\displaystyle y \le 700
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{First we will convert} \text{these inequalities into equations as follows:}
\displaystyle x + y = 800
\displaystyle 2x + y = 1000
\displaystyle x = 400
\displaystyle y = 700
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,700),\ (200,600),\ (400,200),\ (400,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z = 2x + 1.5y \\ \hline (0,700) & 1050 \\ \hline (200,600) & 1300 \\ \hline (400,200) & 1100 \\ \hline (400,0) & 800 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 1300 \text{ which is attained at } (200,600).
\displaystyle \text{Thus, the maximum profit obtained is Rs } 1300 \text{ when } 200 \text{ belts of type A and } 600 \text{ belts of} \\ \text{type B are produced.}

Question 6.
A small manufacturer has employed 5 skilled men and 10 semi-skilled men and makes an article in two varieties, deluxe model and ordinary model. The making of a deluxe model requires 2 hrs work by a skilled man and 2 hrs work by a semi-skilled man. The ordinary model requires 1 hr by a skilled man and 3 hrs by a semi-skilled man. The union rules no man may work more than 8 hrs per day. The manufacturers clear profit on deluxe model is Rs 15 and on an ordinary model is Rs 10. How many of each type should be made in order to maximize his total daily profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ articles of deluxe model and } y \text{ articles of an ordinary model be made.}
\displaystyle \text{Numbers cannot be negative.}
\displaystyle \text{Therefore,}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{According to the question, the profit on each model of deluxe and ordinary type model} \\ \text{are Rs } 15 \text{ and Rs } 10 \text{ respectively.}
\displaystyle \text{So, profits on } x \text{ deluxe model and ordinary models are } 15x \text{ and } 10y.
\displaystyle \text{Let } Z \text{ be total profit, then,}
\displaystyle Z = 15x + 10y
\displaystyle \text{Since, the making of a deluxe and ordinary model requires } 2 \text{ hrs and }  1 \text{ hr work by} \\ \text{skilled men, so } x \text{ deluxe and } y \text{ ordinary models require } 2x \text{ and } y \text{ hours of skilled} \\ \text{men but time available by skilled men is } 5 \times 8 = 40 \text{ hours.}
\displaystyle \text{So,}
\displaystyle 2x + y \le 40
\displaystyle \text{Since, the making of a deluxe and ordinary model requires } 2 \text{ hrs and } 3 \text{ hrs work by} \\ \text{semi-skilled men, so } x \text{ deluxe and } y \text{ ordinary models require } 2x \text{ and } 3y \text{ hours}  \text{of} \\ \text{semi-skilled men but time available by semi-skilled men is } 10 \times 8 = 80 \text{ hours.}
\displaystyle \text{So,}
\displaystyle 2x + 3y \le 80
\displaystyle \text{Hence the mathematical} \\ \text{formulation of LPP is,}
\displaystyle \text{Maximize } Z = 15x + 10y
\displaystyle \text{Subject to:}
\displaystyle 2x + y \le 40
\displaystyle 2x + 3y \le 80
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (20,0),\ (10,20),\ \!\left(0,\frac{80}{3}\right).
\displaystyle \text{The value of } Z = 15x + 10y \text{ at these corner points are}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z = 15x + 10y \\ \hline (20,0) & 300 \\ \hline (10,20) & 350 \\ \hline \!\left(0,\frac{80}{3}\right) & \frac{800}{3} \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 350 \text{ which is attained at } (10,20).
\displaystyle \text{Thus, maximum profit is obtained when } 10 \text{ units of deluxe model and } 20 \text{ units of} \\ \text{ordinary model is produced.}

Question 7.
A manufacturer makes two types of A and B tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Item} & \text{Machine I} & \text{Machine II} & \text{Machine III} \\ \hline A & 12 & 18 & 6 \\ \hline B & 6 & 0 & 9 \\ \hline \end{array}

Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit. (CBSE 2003, 2008)

\displaystyle \text{Answer:}
\displaystyle \text{Let the required number of tea cups of Type A and B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, the profit on each cup A is } 75 \text{ paise and that on each cup B is } 50 \text{ paise. So,} \\ \text{the profit on } x \text{ tea cups of type A and } y \text{ tea cups of type B are } 75x \text{ and } 50y \text{ respectively.}
\displaystyle \text{Let total profit on tea cups be } Z,
\displaystyle Z = 75x + 50y
\displaystyle \text{Since, each tea cup of type A and B require to work machine I for } 12 \text{ and } 6 \text{ minutes} \\ \text{respectively so } x \text{ tea cups of Type A and } y \text{ tea cups of Type B require to work on} \\ \text{machine I for } 12x \text{ and } 6y \text{ minutes respectively.}
\displaystyle \text{Total time available on machine I is } 6 \times 60 = 360 \text{ minutes. So,}
\displaystyle 12x + 6y \le 360
\displaystyle \text{Since, each tea cup of type A and B require to work machine II for } 18 \text{ and } 0 \text{ minutes} \\ \text{respectively so } x \text{ tea cups of Type A and } y \text{ tea cups of Type B require to work on} \\ \text{machine II for } 18x \text{ and } 0y \text{ minutes respectively.}
\displaystyle \text{Total time available on machine II is } 6 \times 60 = 360 \text{ minutes. So,}
\displaystyle 18x + 0y \le 360
\displaystyle x \le 20
\displaystyle \text{Since, each tea cup of type A and B require to work machine III for } 6 \text{ and } 9 \text{ minutes} \\ \text{respectively so } x \text{ tea cups of Type A and } y \text{ tea cups of Type B require to work on} \\ \text{machine III for } 6x \text{ and } 9y \text{ minutes respectively.}
\displaystyle \text{Total time available on machine III is } 6 \times 60 = 360 \text{ minutes. So,}
\displaystyle 6x + 9y \le 360
\displaystyle \text{Hence mathematical formulation of LPP is,}
\displaystyle \text{Maximize } Z = 75x + 50y
\displaystyle \text{Subject to:}
\displaystyle 12x + 6y \le 360
\displaystyle x \le 20
\displaystyle 6x + 9y \le 360
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The shaded region is the feasible region determined by the constraints}
\displaystyle 12x + 6y \le 360
\displaystyle x \le 20
\displaystyle 6x + 9y \le 360
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The corner points are } (0,40),\ (15,30),\ (20,20),\ (20,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z = 75x + 50y \\ \hline (0,40) & 2000 \\ \hline (15,30) & 2625 \\ \hline (20,20) & 2500 \\ \hline (20,0) & 1500 \\ \hline \end{array}
\displaystyle \text{Here } Z \text{ is maximum at } (15,30).
\displaystyle \text{Therefore, } 15 \text{ tea cups of Type A and } 30 \text{ tea cups of Type B are needed to maximize the profit.}

Question 8.
A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Machine} & \text{Area (sq m)} & \text{Labour (men)} & \text{Daily Output} \\ \hline A & 1000 & 12 & 60 \\ B & 1200 & 8 & 40 \\ \hline \end{array}

He has an area of 7600 sq. m available and 72 skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output? (CBSE 2003, 2008) 

\displaystyle \text{Answer:}
\displaystyle \text{Let required number of machine A and B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, products of each machine A and B are } 60 \text{ and } 40 \text{ units daily respectively. So,} \\ \text{produced by } x \text{ number of machine A and } y \text{ number of machine B are } 60x \text{ and } \\ 40y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denotes total output daily, so,}
\displaystyle Z = 60x + 40y
\displaystyle \text{Since, each machine of type A and B requires } 1000 \text{ sq. m and } 1200 \text{ sq. m area so, } \\ x \text{ machine of type A and } y \text{ machine of type B require } 1000x \text{ and } 1200y \text{ sq. m area} \\ \text{but total available area for machine is } 7600 \text{ sq. m. So,}
\displaystyle 1000x + 1200y \le 7600
\displaystyle \text{or,}
\displaystyle 5x + 6y \le 38
\displaystyle \text{Since each machine of type A and B requires } 12 \text{ men and } 8 \text{ men to work respectively.} \\ \text{So, } x \text{ machine of type A and } y \text{ machine of type B require } 12x \text{ and } 8y  \\ \text{ men to work respectively.}
\displaystyle \text{But total men available for work are } 72.
\displaystyle \text{So,}
\displaystyle 12x + 8y \le 72
\displaystyle \text{or,}
\displaystyle 3x + 2y \le 18
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 60x + 40y \text{Subject to:}
\displaystyle 5x + 6y \le 38
\displaystyle 3x + 2y \le 18
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{Shaded region represents the feasible region.}
\displaystyle \text{The corner points are } (0,0),\ \!\left(0,\frac{19}{3}\right),\ (4,3),\ (6,0).
\displaystyle \text{Thus the values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 60x + 40y \\ \hline (0,0) & 0 \\ \hline \left(0,\frac{19}{3}\right) & \frac{760}{3} \\ \hline (4,3) & 360 \\ \hline (6,0) & 360 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 360 \text{ which is attained at } (4,3) \text{ and } (6,0).
\displaystyle \text{Thus, the maximum output is } 360 \text{ units obtained when } 4 \text{ units of type A and } \\ 3 \text{ units of type B or } 6 \text{ units of type A and } 0 \text{ units of type B are manufactured.}

Question 9.
A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let required number of goods A and B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, profits of each A and B are Rs } 40 \text{ and Rs } 50 \text{ respectively. So, profits on } \\ x \text{ number of type A and } y \text{ number of type B are } 40x \text{ and } 50y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote total output daily, so,}
\displaystyle Z = 40x + 50y
\displaystyle \text{Since, each A and B requires } 3 \text{ grams and } 1 \text{ gram of silver respectively. So, } \\ x \text{ of type A and } y \text{ of type B require } 3x \text{ and } y \text{ of silver respectively.}
\displaystyle \text{But total silver available is } 9 \text{ grams. So,}
\displaystyle 3x + y \le 9
\displaystyle \text{Since each A and B requires } 1 \text{ gram and } 2 \text{ grams of gold respectively. So, } \\ x \text{ of type A and } y \text{ of type B require } x \text{ and } 2y \text{ of gold respectively.}
\displaystyle \text{But total gold available is } 8 \text{ grams.}
\displaystyle \text{So,}
\displaystyle x + 2y \le 8
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 40x + 50y
\displaystyle \text{Subject to:}
\displaystyle 3x + y \le 9
\displaystyle x + 2y \le 8
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,4),\ (2,3),\ (3,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 40x + 50y \\ \hline (0, 0) & 0 \\ \hline (0, 4) & 200 \\ \hline (2, 3) & 230 \\ \hline (3, 0) & 120 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 230 \text{ which is attained at } (2,3).
\displaystyle \text{Thus the maximum profit is of Rs } 230 \text{ when } 2 \text{ units of Type A and } 3 \text{ units of Type B are produced.}

Question 10.
A manufacturer of furniture makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.

\displaystyle \text{Answer:}
\displaystyle \text{Let daily production of chairs and tables be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, profits of each chair and table are Rs } 3 \text{ and Rs } 5 \text{ respectively. So, profits on } \\ x \text{ number of chairs and } y \text{ number of tables are } 3x \text{ and } 5y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote total output daily, so,}
\displaystyle Z = 3x + 5y
\displaystyle \text{Since, each chair and table requires } 2 \text{ hrs and } 3 \text{ hrs on machine A respectively. So, } \\ x \text{ chairs and } y \text{ tables require } 2x \text{ and } 3y \text{ hrs on machine A respectively.}
\displaystyle \text{Total time available on Machine A is } 16 \text{ hours. So,}
\displaystyle 2x + 3y \le 16
\displaystyle \text{Dividing by } 2,
\displaystyle x + \frac{3}{2}y \le 8
\displaystyle \text{Since, each chair and table require } 6 \text{ hrs and } 2 \text{ hrs on machine B respectively. So, } \\ x \text{ chairs and } y \text{ tables require } 6x \text{ and } 2y \text{ hrs on machine B respectively.}
\displaystyle \text{Total time available on Machine B is } 30 \text{ hours. So,}
\displaystyle 6x + 2y \le 30
\displaystyle \text{Dividing by } 2,
\displaystyle 3x + y \le 15
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 3x + 5y
\displaystyle \text{Subject to:}
\displaystyle x + \frac{3}{2}y \le 8
\displaystyle 3x + y \le 15
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,4),\ \!\left(\frac{22}{5},\frac{9}{5}\right),\ (5,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 3x + 5y \\ \hline (0, 0) & 0 \\ \hline (0, 4) & 20 \\ \hline \!\left(\frac{22}{5},\frac{9}{5}\right) & \frac{111}{5} \\ \hline (5, 0) & 15 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } \frac{111}{5} \text{ which is attained at } \!\left(\frac{22}{5},\frac{9}{5}\right).
\displaystyle \text{Thus the maximum profit is Rs } \frac{111}{5} \text{ when } \frac{22}{5} \text{ units of chair and } \frac{9}{5} \text{ units of table are produced.}

Question 11.
A furniture manufacturing company plans to make two products: chairs and tables. From its available resources which consists of 400 square feet of teak wood and 450 man hours. It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs 80. How many items of each product should be produced by the company so that the profit is maximum?

\displaystyle \text{Answer:}
\displaystyle \text{Let required production of chairs and tables be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, profits of each chair and table are Rs } 45 \text{ and Rs } 80 \text{ respectively. So,} \\ \text{profits on } x \text{ number of chairs and } y \text{ number of tables are } 45x \text{ and } 80y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote total output daily, so,}
\displaystyle Z = 45x + 80y
\displaystyle \text{Since, each chair and table requires } 5 \text{ sq.\ ft and } 80 \text{ sq.\ ft of wood} \\ \text{respectively. So, } x \text{ chairs and } y \text{ tables require } 5x \text{ and } 80y \text{ sq.\ ft of wood respectively.}
\displaystyle \text{But } 400 \text{ sq.\ ft of wood is available. So,}
\displaystyle 5x + 80y \le 400
\displaystyle \text{Dividing by } 5,
\displaystyle x + 4y \le 80
\displaystyle \text{Since, each chair and table requires } 10 \text{ and } 25 \text{ men-hours respectively. So, }  \\ x \text{ chairs and } y \text{ tables require } 10x \text{ and } 25y \text{ men-hours respectively.}
\displaystyle \text{But only } 450 \text{ hours are available. So,}
\displaystyle 10x + 25y \le 450
\displaystyle \text{Dividing by } 5,
\displaystyle 2x + 5y \le 90
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 45x + 80y
\displaystyle \text{Subject to:}
\displaystyle x + 4y \le 80
\displaystyle 2x + 5y \le 90
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,18),\ (45,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 45x + 80y \\ \hline (0, 0) & 0 \\ \hline (0, 18) & 1440 \\ \hline (45,0) & 2025 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 2025 \text{ which is attained at } (45,0).
\displaystyle \text{Thus maximum profit of Rs } 2025 \text{ is obtained when } 45 \text{ units of chairs and } \\ 0 \text{ units of tables are produced.}

Question 12.
A firm manufactures two products A and B. Each product is processed on two machines M₁ and M₂. Product A requires 4 minutes of processing time on M₁ and 8 min. on M₂; product B requires 4 minutes on M₁ and 4 min. on M₂. The machine M₁ is available for not more than 8 hrs 20 min while machine M₂ is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively. Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.

\displaystyle \text{Answer:}
\displaystyle \text{Let required production of product A and B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since profit on each product A and B are Rs } 3 \text{ and Rs } 4 \text{ respectively. So,} \\ \text{profits on } x \text{ number of type A and } y \text{ number of type B are } 3x \text{ and } 4y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote total output daily, so,}
\displaystyle Z = 3x + 4y
\displaystyle \text{Since, each A and B requires } 4 \text{ minutes each on machine } M_1. \text{ So, } \\ x \text{ of type A and } y \text{ of type B require } 4x \text{ and } 4y \text{ minutes respectively.}
\displaystyle \text{Total time available on machine } M_1 \text{ is } 8 \text{ hours } 20 \text{ minutes } = 500 \text{ minutes.}
\displaystyle \text{So,}
\displaystyle 4x + 4y \le 500
\displaystyle \text{Dividing by } 4,
\displaystyle x + y \le 125
\displaystyle \text{Since, each A and B requires } 8 \text{ minutes and } 4 \text{ minutes on machine } \\ M_2 \text{ respectively. So, } x \text{ of type A and } y \text{ of type B require } 8x \text{ and } 4y \text{ minutes respectively.}
\displaystyle \text{Total time available on machine } M_2 \text{ is } 10 \text{ hours } = 600 \text{ minutes.}
\displaystyle \text{So,}
\displaystyle 8x + 4y \le 600
\displaystyle \text{Dividing by } 4,
\displaystyle 2x + y \le 150
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 3x + 4y
\displaystyle \text{Subject to:}
\displaystyle x + y \le 125
\displaystyle 2x + y \le 150
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } O(0,0),\ B(0,125),\ E(25,100),\ C(75,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 3x + 4y \\ \hline (0,0) & 0 \\ \hline (0,125) & 500 \\ \hline (25,100) & 475 \\ \hline (75,0) & 225 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 500 \text{ which is attained at } (0,125).
\displaystyle \text{Thus, the maximum profit is Rs } 500 \text{ obtained when } 0 \text{ units of product A and } \\  125 \text{ units of product B are manufactured.}

Question 13.
A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programming problem for maximum profit and solve it graphically.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of item A and } y \text{ units of item B be manufactured.}
\displaystyle \text{Numbers of items cannot be negative.}
\displaystyle \text{Therefore,}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The given information can be tabulated as follows:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Product} & \text{Motors} & \text{Transformers} \\ \hline A(x) & 3 & 4 \\ \hline B(y) & 2 & 4 \\ \hline \text{Availability} & 210 & 300 \\ \hline \end{array}
\displaystyle \text{Further, it is given that type B is an export model, whose supply is restricted to }  \\ 65 \text{ units per month.}
\displaystyle \text{Therefore, the constraints are}
\displaystyle 3x + 2y \le 210
\displaystyle 4x + 4y \le 300
\displaystyle y \le 65
\displaystyle \text{A and B can make profit of Rs } 20 \text{ and Rs } 30 \text{ per unit respectively.}
\displaystyle \text{Therefore, profit gained from } x \text{ units of item A and } y \text{ units of item B is Rs } 20x  \\ \text{ and } 30y \text{ respectively.}
\displaystyle \text{Total profit } Z = 20x + 30y \text{ which according to the question is to be maximized.}
\displaystyle \text{Thus the mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 20x + 30y
\displaystyle \text{Subject to:}
\displaystyle 3x + 2y \le 210
\displaystyle 4x + 4y \le 300
\displaystyle y \le 65
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,65),\ (10,65),\ (60,15),\ (70,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 20x + 30y \\ \hline (0,0) & 0 \\ \hline (0,65) & 1950 \\ \hline (10,65) & 2150 \\ \hline (60,15) & 1650 \\ \hline (70,0) & 1400 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 2150 \text{ which is attained at } (10,65).
\displaystyle \text{Thus, the maximum profit is Rs } 2150 \text{ obtained when } 10 \text{ units of item A and } \\ 65 \text{ units of item B are manufactured.}

Question 14.
A factory uses three different resources for the manufacture of two different products. 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second product 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.

\displaystyle \text{Answer:}
\displaystyle \text{Let number of product I and product II be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since, profits on each product I and II are } 2 \text{ and } 3 \text{ monetary units. So, profits on } \\ x \text{ number of product I and } y \text{ number of product II are } 2x \text{ and } 3y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote total output daily, so,}
\displaystyle Z = 2x + 3y
\displaystyle \text{Since, each I and II requires } 2 \text{ and } 4 \text{ units of resource A. So, } x \text{ units of product I} \\ \text{and } y \text{ units of product II require } 2x \text{ and } 4y \text{ units respectively. But maximum} \\ \text{available quantity of resource A is } 20 \text{ units.}
\displaystyle \text{So,}
\displaystyle 2x + 4y \le 20
\displaystyle \text{Dividing by } 2,
\displaystyle x + 2y \le 10
\displaystyle \text{Since, each I and II requires } 2 \text{ and } 2 \text{ units of resource B. So, } x \text{ units of product I} \\ \text{and } y \text{ units of product II require } 2x \text{ and } 2y \text{ units respectively. But maximum} \\ \text{available quantity of resource B is } 12 \text{ units.}
\displaystyle \text{So,}
\displaystyle 2x + 2y \le 12
\displaystyle \text{Dividing by } 2,
\displaystyle x + y \le 6
\displaystyle \text{Since, each unit of product I requires } 4 \text{ units of resource C. It is not required by product II.} \\ \text{So, } x \text{ units of product I require } 4x \text{ units of resource C. But maximum} \\ \text{available quantity of resource C is } 16 \text{ units.}
\displaystyle \text{So,}
\displaystyle 4x \le 16
\displaystyle x \le 4
\displaystyle \text{Hence mathematical formulation of LPP is,}
\displaystyle \text{Maximize } Z = 2x + 3y
\displaystyle \text{Subject to:}
\displaystyle x + 2y \le 10
\displaystyle x + y \le 6
\displaystyle x \le 4
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,5),\ (2,4),\ (4,2),\ (4,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 2x + 3y \\ \hline (0,0) & 0 \\ \hline (0,5) & 15 \\ \hline (2,4) & 16 \\ \hline (4,2) & 14 \\ \hline (4,0) & 8 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 16 \text{ which is attained at } (2,4).
\displaystyle \text{Thus, the maximum profit is } 16 \text{ monetary units obtained when } 2 \text{ units of first} \\ \text{product and } 4 \text{ units of second product were manufactured.}

Question 15.
A publisher sells a hard cover edition of a text book for Rs 72.00 and a paperback edition of the same text for Rs 40.00. Costs to the publisher are Rs 56.00 and Rs 28.00 per book respectively in addition to weekly costs of Rs 9600.00. Both books require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations are 4800 minutes available each week. How many of each type of book should be produced in order to maximize profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let the sale of hard cover edition be } h \text{ and that of paperback edition be } t.
\displaystyle \text{SP of a hard cover edition of the textbook = Rs } 72.
\displaystyle \text{SP of a paperback edition of the textbook = Rs } 40.
\displaystyle \text{Cost to the publisher for hard cover edition = Rs } 56.
\displaystyle \text{Cost to the publisher for a paperback edition = Rs } 28.
\displaystyle \text{Weekly cost to the publisher = Rs } 9600.
\displaystyle \text{Profit to be maximized } Z = (72-56)h + (40-28)t - 9600.
\displaystyle Z = 16h + 12t - 9600
\displaystyle \text{Suppose printing cost per hard cover copy is Rs } 5 \text{ and per paperback copy is Rs } 2.
\displaystyle 5(h+t) \le 4800
\displaystyle 10h + 2t \le 4800
\displaystyle \text{The corner points are } O(0,0),\ B_1(0,960),\ E_1(360,600),\ F_1(480,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 16h + 12t - 9600 \\ \hline O(0,0) & -9600 \\ \hline B_1(0,960) & 1920 \\ \hline E_1(360,600) & 3360 \\ \hline F_1(480,0) & -1920 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 3360 \text{ which is attained at } E_1(360,600).
\displaystyle \text{The maximum profit is } 3360 \text{ which is obtained by selling } 360 \text{ copies of hard} \\ \text{cover edition and } 600 \text{ copies of paperback edition.}

Question 16.
A firm manufactures headache pills of two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effect. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.

\displaystyle \text{Answer:}
\displaystyle \textbf{Let }x=\text{number of size A pills, and }y=\text{number of size B pills.}
\displaystyle \textbf{Minimize (least pills):}\quad N=x+y.
\displaystyle \textbf{Subject to:}
\displaystyle \text{Aspirin:}\quad 2x+y\ge 12.
\displaystyle \text{Bicarbonate:}\quad 5x+8y\ge 7.4.
\displaystyle \text{Codeine:}\quad x+66y\ge 24.
\displaystyle x\ge 0,\;y\ge 0.\displaystyle \textbf{Since }5x+8y\ge 7.4\text{ is very small compared to the other needs, the minimum pills} \\ \text{will occur at the intersection of the tighter constraints }2x+y=12\text{ and }x+66y=24.
\displaystyle \textbf{Solve }2x+y=12\text{ and }x+66y=24:
\displaystyle y=12-2x.
\displaystyle x+66(12-2x)=24.
\displaystyle x+792-132x=24.
\displaystyle -131x=-768\Rightarrow x=\frac{768}{131}.
\displaystyle y=12-2\left(\frac{768}{131}\right)=\frac{36}{131}.
\displaystyle \textbf{Check bicarbonate: }5x+8y=5\left(\frac{768}{131}\right)+8\left(\frac{36}{131}\right)=\frac{4128}{131}\ge 7.4\;\Rightarrow\;\text{feasible.}
\displaystyle \textbf{Minimum number of pills: }N_{\min}=x+y=\frac{768}{131}+\frac{36}{131}=\frac{804}{131}\approx 6.14.
\displaystyle \textbf{Quantity of codeine consumed at optimum: }x+66y=24\text{ grains.}

Question 17.
A chemical company produces two compounds, A and B. The following table gives the units of ingredients C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Ingredient} & A & B & \text{Minimum Requirement} \\ \hline C & 1 & 2 & 80 \\ D & 3 & 1 & 75 \\ \text{Cost (Rs/kg)} & 4 & 6 & {} \\ \hline \end{array}

\displaystyle \text{Answer:}
\displaystyle \text{Let required quantity of compound A and B be } x \text{ and } y \text{ kg.}
\displaystyle \text{Since, cost of one kg of compound A and B are Rs } 4 \text{ and Rs } 6 \text{ per kg. So, cost of }  x \text{ kg of} \\ \text{compound A and } y \text{ kg of compound B are Rs } 4x \text{ and Rs } 6y \text{ respectively.}
\displaystyle \text{Let } Z \text{ be the total cost of compounds, so,}
\displaystyle Z = 4x + 6y
\displaystyle \text{Since, compound A and B contain } 1 \text{ and } 2 \text{ units of ingredient C per kgrespectively. So } x \text{ kg of} \\ \text{compound A and } y \text{ kg of compound B contain } x \text{ and } 2y \text{ units of ingredient C respectively but} \\ \text{minimum requirement of ingredient C is } 80 \text{ units, so,}
\displaystyle x + 2y \ge 80
\displaystyle \text{Since, compound A and B contain } 3 \text{ and } 1 \text{ units of ingredient D per kg respectively. So } x \text{ kg of} \\ \text{compound A and } y \text{ kg of compound B contain } 3x \text{ and } y \text{ units of ingredient D respectively but} \\ \text{minimum requirement of ingredient D is } 75 \text{ units, so,}
\displaystyle 3x + y \ge 75
\displaystyle \text{Hence mathematical formulation of LPP is,}
\displaystyle \text{Minimize } Z = 4x + 6y
\displaystyle \text{Subject to:}
\displaystyle x + 2y \ge 80
\displaystyle 3x + y \ge 75
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,75),\ (14,33),\ (80,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \text{Corner Point } (0,75) \Rightarrow 450
\displaystyle \text{Corner Point } (14,33) \Rightarrow 254
\displaystyle \text{Corner Point } (80,0) \Rightarrow 320
\displaystyle \text{The minimum value of } Z \text{ is } 254 \text{ which is attained at } (14,33).
\displaystyle \text{Thus, the minimum cost is Rs } 254 \text{ obtained when } 14 \text{ units of compound A and } \\ 33 \text{ units of compound B are produced.}

Question 18.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let the company manufacture } x \text{ souvenirs of Type A and } y \text{ souvenirs of Type B.}
\displaystyle \text{Therefore, } x \ge 0,\ y \ge 0
\displaystyle \text{The given information can be compiled in a table as follows:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Operation} & \text{Type A} & \text{Type B} & \text{Availability} \\ \hline \text{Cutting (min)} & 5 & 8 & 3 \times 60 + 20 = 200 \\ \hline \text{Assembling (min)} & 10 & 8 & 4 \times 60 = 240 \\ \hline \end{array}
\displaystyle \text{The profit on Type A souvenirs is } 50 \text{ paisa and on Type B souvenirs is } 60 \text{ paisa.}
\displaystyle \text{Therefore, profit gained on souvenirs of Type A and Type B is Rs } 0.50x \text{ and Rs } 0.60y \text{ respectively.}
\displaystyle \text{Total profit } Z = 0.5x + 0.6y
\displaystyle \text{The mathematical formulation of the given problem is,}
\displaystyle \text{Maximize } Z = 0.5x + 0.6y
\displaystyle \text{Subject to:}
\displaystyle 5x + 8y \le 200
\displaystyle 10x + 8y \le 240
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}

\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,25),\ (8,20),\ (24,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 0.5x + 0.6y \\ \hline (0, 0) & 0 \\ \hline (0,25) & 15 \\ \hline (8,20) & 16 \\ \hline (24, 0) & 12 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is attained at } (8,20).
\displaystyle \text{Thus, } 8 \text{ souvenirs of Type A and } 20 \text{ souvenirs of Type B should be produced each day to} \\ \text{get the maximum profit of Rs } 16.

Question 19.
A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of product A and } y \text{ units of product B be manufactured.}
\displaystyle \text{Number of units cannot be negative.}
\displaystyle \text{Therefore, } x \ge 0,\ y \ge 0.
\displaystyle \text{According to the question, the given information can be tabulated as:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Product} & \text{Selling price (Rs)} & \text{Manufacturing time (hrs)} \\ \hline \text{Product A } (x) & 200 & 0.5 \\ \hline \text{Product B } (y) & 300 & 1 \\ \hline \end{array}
\displaystyle \text{Also, the availability of time is } 40 \text{ hours and the revenue should be at least Rs } 10000.
\displaystyle \text{Further, it is given that there is a permanent order for } 14 \text{ units of Product A and } \\ 16 \text{ units of Product B.}
\displaystyle \text{Therefore, the constraints are,}
\displaystyle 200x + 300y \ge 10000
\displaystyle 0.5x + y \le 40
\displaystyle x \ge 14
\displaystyle y \ge 16
\displaystyle \text{If the profit on each of product A is Rs } 20 \text{ and on product B is Rs } 30.
\displaystyle \text{Therefore, profit gained on } x \text{ units of product A and } y \text{ units of product B is Rs } \\ 20x \text{ and Rs } 30y.
\displaystyle \text{Total profit } Z = 20x + 30y \text{ which is to be maximized.}
\displaystyle \text{Thus, the mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 20x + 30y
\displaystyle \text{Subject to:}
\displaystyle 200x + 300y \ge 10000
\displaystyle 0.5x + y \le 40
\displaystyle x \ge 14
\displaystyle y \ge 16
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (26,16),\ (48,16),\ (14,33),\ (14,24).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 20x + 30y \\ \hline (26,16) & 1000 \\ \hline (48,16) & 1440 \\ \hline (14,33) & 1270 \\ \hline (14,24) & 1000 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is Rs } 1440 \text{ which is attained at } (48,16).
\displaystyle \text{Thus, the maximum profit is Rs } 1440 \text{ obtained when } 48 \text{ units of product A and } \\ 16 \text{ units of product B are manufactured.}

Question 20.
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first type of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first and second type respectively. How many trunks of each type must he make per day to maximize profit? (CBSE 2001, 2005, 2012)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ trunks of first type and } y \text{ trunks of second type be manufactured.}
\displaystyle \text{Number of trunks cannot be negative.} \text{Therefore, } x \ge 0,\ y \ge 0
\displaystyle \text{According to the question, the given information can be tabulated as}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Type} & \text{Machine A (hours)} & \text{Machine B (hours)} \\ \hline \text{First type } (x) & 3 & 3 \\ \hline \text{Second type } (y) & 3 & 2 \\ \hline \text{Availability} & 18 & 15 \\ \hline \end{array}
\displaystyle \text{Therefore, the constraints are,}
\displaystyle 3x + 3y \le 18
\displaystyle 3x + 2y \le 15
\displaystyle \text{He earns a profit of Rs } 30 \text{ and Rs } 25 \text{ per trunk of the first type and the second type respectively.}
\displaystyle \text{Therefore, profit gained by him from } x \text{ trunks of first type and } y \text{ trunks of second} \\ \text{type is Rs } 30x \text{ and Rs } 25y \text{ respectively.}
\displaystyle \text{Total profit } Z = 30x + 25y \text{ which is to be maximized.}
\displaystyle \text{Thus, the mathematical formulation of the given LPP is}
\displaystyle \text{Maximize } Z = 30x + 25y
\displaystyle \text{Subject to:}
\displaystyle 3x + 3y \le 18
\displaystyle 3x + 2y \le 15
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,6),\ (3,3),\ s(5,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 30x + 25y \\ \hline (0,0) & 0 \\ \hline (0,6) & 150 \\ \hline (3,3) & 165 \\ \hline (5,0) & 150 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 165 \text{ which is attained at } E(3,3).
\displaystyle \text{Thus, the maximum profit of Rs } 165 \text{ is obtained when } 3 \text{ units of each type of trunk} \\ \text{are manufactured.}

Question 21.
A manufacturer of patent medicines is preparing a product of three medicines, A and B. There are sufficient raw materials available to make 20000 bottles of A and 40000 bottles of B, but there are only 45000 bottles into which the medicines can be poured. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of A; it takes 1 hour to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation. The profit is Rs 8 per bottle for A and Rs 7 per bottle for B. How should the manufacturer schedule his production in order to maximize his profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let production of each bottle of A and B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Since profits on each bottle of A and B are Rs } 8 \text{ and Rs } 7 \text{ per bottle respectively,} \\ \text{profit on } x \text{ bottles of A and } y \text{ bottles of B are } 8x \text{ and } 7y \text{ respectively.}
\displaystyle \text{Let } Z \text{ be total profit on bottles so,}
\displaystyle Z = 8x + 7y
\displaystyle \text{Since it takes } 3 \text{ hours and } 1 \text{ hour to prepare enough material to fill } 1000 \\ \text{ bottles of Type A and Type B respectively, so } x \text{ bottles of A and } y \text{ bottles of B require } \\ \frac{3x}{1000} \text{ hours} \text{and } \frac{y}{1000} \text{ hours respectively, but only } 66 \text{ hours are available, so,}
\displaystyle \frac{3x}{1000} + \frac{y}{1000} \le 66
\displaystyle 3x + y \le 66000
\displaystyle \text{Since raw materials available can make } 20000 \text{ bottles of A and } 40000 \text{ bottles of} \\ \text{B but there are } 45000 \text{ bottles in which either of these medicines can be put, so,}
\displaystyle x \le 20000
\displaystyle y \le 40000
\displaystyle x + y \le 45000
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{Since production of bottles cannot be negative.}
\displaystyle \text{Hence mathematical formulation of the given LPP is,}
\displaystyle \text{Maximize } Z = 8x + 7y
\displaystyle \text{Subject to:}
\displaystyle 3x + y \le 66000
\displaystyle x \le 20000
\displaystyle y \le 40000
\displaystyle x + y \le 45000
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,0),\ (0,40000),\ (10500,34500),\ (20000,6000),\ (20000,0).
\displaystyle \text{The values of } Z \text{ at these corner points are,}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 8x + 7y \\ \hline (0, 0) & 0 \\ \hline (0,40000) & 280000 \\ \hline (10500,34500) & 325500 \\ \hline (20000,6000) & 188000 \\ \hline (20000,0) & 160000 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 325500 \text{ which is attained at } (10500,34500).
\displaystyle \text{Thus the maximum profit is Rs } 325500 \text{ obtained when } 10500 \text{ bottles of A and } \\ 34500 \text{ bottles of B are manufactured.}

Question 22.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to first class. Determine how many each type of tickets must be sold in order to maximize profit for the airline. What is the maximum profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let required number of first class and economy class tickets be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Each ticket of first class and economy class makes profit of Rs } 400 \text{ and Rs } 600 \text{ respectively.}
\displaystyle \text{So, } x \text{ tickets of first class and } y \text{ tickets of economy class make profit of Rs } 400x \\ \text{and Rs } 600y \text{ respectively.}
\displaystyle \text{Let total profit be } Z = 400x + 600y
\displaystyle \text{Given, aeroplane can carry a minimum of } 200 \text{ passengers, Hence, } x + y \le 200
\displaystyle \text{Given airline reserves at least } 20 \text{ seats for first class, Hence, } x \ge 20
\displaystyle \text{Also, at least } 4 \text{ times as many passengers prefer to travel by economy class as first class, so,} y \ge 4x
\displaystyle \text{Hence the mathematical formulation of the LPP is}
\displaystyle \text{Maximize } Z = 400x + 600y
\displaystyle \text{Subject to:}
\displaystyle x + y \le 200
\displaystyle y \ge 4x
\displaystyle x \ge 20
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (20,80),\ (40,160),\ (20,180).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 400x + 600y \\ \hline (0, 0) & 0 \\ \hline (20,80) & 56000 \\ \hline (40,160) & 112000 \\ \hline (20,180) & 116000 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is attained at } (20,180).
\displaystyle \text{Thus, the maximum profit is Rs } 116000 \text{ obtained when } 20 \text{ first class tickets and } 180 \\ \text{ economy class tickets are sold.}

Question 23.
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at minimum cost. What is the minimum cost? (CBSE 2002, 2008)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ kg of Type I fertilizer and } y \text{ kg of Type II fertilizers be supplied.}
\displaystyle \text{The quantity of fertilizers cannot be negative.} \text{So, } x \ge 0,\ y \ge 0
\displaystyle \text{A gardener has a supply of fertilizer of Type I which consists of } 10\% \text{ nitrogen and } \\ \text{Type II consists of } 5\% \text{ nitrogen, and he needs at least } 14 \text{ kg of nitrogen for his crop.}
\displaystyle \text{So, } 0.10x + 0.05y \ge 14 \text{ Multiplying throughout by } 100, \text{ we get } \\ 10x + 5y \ge 1400
\displaystyle \text{A gardener has a supply of fertilizer of Type I which consists of } 6\% \text{ phosphoric acid and} \\ \text{Type II consists of } 10\% \text{ phosphoric acid, and he needs at least } 14 \text{ kg of phosphoric acid} \\ \text{for his crop.}
\displaystyle \text{So, } 0.06x + 0.10y \ge 14 \text{ Multiplying throughout by } 100, \text{ we get } \\ 6x + 10y \ge 1400
\displaystyle \text{Therefore, the constraints are,}
\displaystyle 10x + 5y \ge 1400
\displaystyle 6x + 10y \ge 1400
\displaystyle \text{If Type I fertilizer costs } 60 \text{ paise per kg and Type II fertilizer costs } 40 \text{ paise per kg.}
\displaystyle \text{Therefore, the cost of } x \text{ kg of Type I fertilizer and } y \text{ kg of Type II fertilizer is Rs }\\ 0.60x \text{ and Rs } 0.40y \text{ respectively.}
\displaystyle \text{Total cost } Z = 0.6x + 0.4y \text{ is to be minimized.}
\displaystyle \text{Thus the mathematical formulation of the given LPP is,}
\displaystyle \text{Minimize } Z = 0.6x + 0.4y
\displaystyle \text{Subject to:}
\displaystyle 10x + 5y \ge 1400
\displaystyle 6x + 10y \ge 1400
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,280),\ (100,80),\ \!\left(\frac{700}{3},0\right).
\displaystyle \text{The values of } Z \text{ at these points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 0.6x + 0.4y \\ \hline (0, 0) & 0 \\ \hline (0,280) & 112 \\ \hline (100,80) & 92 \\ \hline \left(\frac{700}{3},0\right) & 140 \\ \hline \end{array}
\displaystyle \text{The minimum value of } Z \text{ is } 92 \text{ which is attained at } (100,80).
\displaystyle \text{Thus, the minimum cost is Rs } 92 \text{ obtained when } 100 \text{ kg of Type I fertilizer and } \\ 80 \text{ kg of Type II fertilizer are supplied.}

Question 24.
Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds. If the rate of interest on saving certificates is 8% per annum and the rate of interest on National Saving Bonds is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income. 

\displaystyle \text{Answer:}
\displaystyle \text{Let Anil invest Rs }x\text{ and Rs }y\text{ in saving certificate (SC) and National saving bond} \\ \text{(NSB) respectively.}
\displaystyle \text{Since, the rate of interest on SC is }8\%\text{ annual and on NSB is }10\%\text{ annual, so} \\ \text{interest on Rs }x\text{ of SC is }\frac{8x}{100}\text{ and Rs }y\text{ of NSB is }\frac{10y}{100}\text{ per annum.}
\displaystyle \text{Let }Z\text{ be total interest earned, so}
\displaystyle Z=\frac{8x}{100}+\frac{10y}{100}
\displaystyle \text{Given he wants to invest Rs }12000\text{ in total}
\displaystyle x+y\le12000
\displaystyle \text{According to the rules he has to invest at least Rs }2000\text{ in SC and at least Rs }4000 \\ \text{ in NSB.}
\displaystyle x\ge2000
\displaystyle y\ge4000
\displaystyle \text{Hence the mathematical formulation of LPP is to find }x\text{ and }y\text{ which maximizes }Z
\displaystyle \text{Max }Z=\frac{8x}{100}+\frac{10y}{100}
\displaystyle \text{Subject to:}
\displaystyle x\ge2000
\displaystyle y\ge4000
\displaystyle x+y\le12000
\displaystyle x,y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are }(2000,10000),(2000,4000),(8000,4000).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z=\frac{8x}{100}+\frac{10y}{100}\\ \hline (0, 0) & 0\\ \hline (2000,10000) & 1160\\ \hline (8000,4000) & 1040\\ \hline (2000,4000) & 560\\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is Rs }1160\text{ which is attained at }(2000,10000).
\displaystyle \text{Thus, the maximum earning is Rs }1160\text{ obtained when Rs }2000\text{ were invested in} \\ \text{SC and Rs }10000\text{ in NSB.}

Question 25.
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit? 

\displaystyle \text{Answer:}

\displaystyle \text{Let the required number of trees of Type A and Type B be } x \text{ and } y \text{ respectively.}
\displaystyle \text{Number of trees cannot be negative. Hence, } x, y \ge 0.
\displaystyle \text{To plant a tree of Type A requires } 10 \text{ sq. m and Type B requires } 20 \text{ sq. m} \\ \text{of ground per tree.}
\displaystyle \text{It is given that a man owns a field of area } 1000 \text{ sq. m. Therefore,}
\displaystyle 10x + 20y \le 1000 \Rightarrow  x + 2y \le 100
\displaystyle \text{Type A costs Rs } 20 \text{ per tree and Type B costs Rs } 25 \text{ per tree.}
\displaystyle \text{Therefore, } x \text{ trees of Type A and } y \text{ trees of Type B cost Rs } 20x \text{ and Rs } 25y \\ \text{ respectively.}
\displaystyle \text{A man has a sum of Rs } 1400 \text{ to purchase young trees.}
\displaystyle 20x + 25y \le 1400
\displaystyle 4x + 5y \le 280
\displaystyle \text{Thus the mathematical formulation of the given LPP is}
\displaystyle \text{Max } Z = 40x - 20x + 60y - 25y = 20x + 35y
\displaystyle \text{Subject to:}
\displaystyle x + 2y \le 100
\displaystyle 4x + 5y \le 280
\displaystyle x, y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are } (70,0), (20,40), (0,50).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Points} & Z = 20x + 35y \\ \hline (0,0) & 0 \\ \hline (70,0) & 1750 \\ \hline (20,40) & 1800 \\ \hline (0,50) & 1400 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 1800 \text{ which is attained at } (20,40).
\displaystyle \text{Thus the maximum profit is Rs } 1800 \text{ obtained when } 20 \text{ trees of Type A and } \\ 40 \text{ trees of Type B are planted.}

Question 26.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is Rs 5.00 and a shade is Rs 3.00. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?  (CBSE 2013)

\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ units of pedestal lamps and } y \text{ units of wooden shades are produced} \\ \text{in a day to maximise the profit.}
\displaystyle \text{Since a pedestal lamp requires } 2 \text{ hours on the grinding/cutting machine and a wooden} \\ \text{shade requires } 1 \text{ hour on the grinding/cutting machine, therefore, the total hours required} \\ \text{for grinding/cutting } x \text{ units of pedestal lamps and } y \text{ units of wooden shades are } (2x+y).
\displaystyle \text{But the grinding/cutting machine is available for at most } 12 \text{ hours on a day.} \\ \therefore\ 2x+y\le 12
\displaystyle \text{Similarly, a pedestal lamp requires } 3 \text{ hours on the sprayer and a wooden shade} \\ \text{requires } 2 \text{ hours on the sprayer, therefore, the total hours required for spraying } \\x \text{ units of pedestallamps and } y \text{ units of wooden shades are } (3x+2y).
\displaystyle \text{But the sprayer is available for at most } 20 \text{ hours on a day.}
\displaystyle \therefore\ 3x+2y\le 20
\displaystyle \text{The profit from the sale of a pedestal lamp is Rs } 5.00 \text{ and a wooden shade is Rs } 3.00.
\displaystyle \text{Therefore, the total profit from the sale of } x \text{ units of pedestal lamps and } y \text{ units} \\ \text{of wooden shades is Rs } (5x+3y).
\displaystyle \text{Thus, the given linear programming problem is}
\displaystyle \text{Maximise } Z=5x+3y
\displaystyle \text{Subject to:}
\displaystyle 2x+y\le 12
\displaystyle 3x+2y\le 20
\displaystyle x\ge 0,\ y\ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The coordinates of the corner points of the feasible region are } (0,0),\ (6,0),\ (4,4)\text{ and } (0,10).
\displaystyle \text{The value of the objective function at these points are given in the following table.}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=5x+3y \\ \hline (0,0) & 5\times 0+3\times 0=0 \\ \hline (6,0) & 5\times 6+3\times 0=30 \\ \hline (4,4) & 5\times 4+3\times 4=32 \\ \hline (0,10) & 5\times 0+3\times 10=30 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 32 \text{ at } x=4,\ y=4.
\displaystyle \text{Hence, the manufacturer should produce } 4 \text{ pedestal lamps and } 4 \text{ wooden shades} \\ \text{to maximise his profit.}
\displaystyle \text{The maximum profit of the manufacturer is Rs } 32 \text{ on a day.}

Question 27.
A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods X and Y. To produce one unit of X, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital are required to produce one unit of Y. If X and Y are priced at Rs 100 and Rs 120 per unit respectively, how should the producer use his resources to maximize total revenue? Solve this problem graphically. (CBSE 2000)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x_1 \text{ and } y_1 \text{ units of goods } x \text{ and } y \text{ be produced respectively.}
\displaystyle \text{Number of units of goods cannot be negative.} \therefore\ x_1,\ y_1\ge 0
\displaystyle \text{To produce one unit of } x,\ 2 \text{ units of labour and for one unit of } y,\ 3 \text{ units of} \\ \text{labour are required. Hence, } 2x_1+3y_1\le 30
\displaystyle \text{To produce one unit of } x,\ 3 \text{ units of capital is required and } 1 \text{ unit of capital is} \\ \text{required to produce one unit of } y. \text{Hence, } 3x_1+y_1\le 17
\displaystyle \text{If } x \text{ and } y \text{ are priced at Rs } 100 \text{ and Rs } 120 \text{ per unit respectively,} \\ \text{therefore, cost of } x_1 \text{ and } y_1 \text{ units of goods } x \text{ and } y \text{ is Rs } 100x_1 \text{ and Rs } 120y_1 \text{ respectively.}
\displaystyle \text{Total revenue } = Z=100x_1+120y_1 \text{ which is to be maximised.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Max } Z=100x_1+120y_1
\displaystyle \text{Subject to:}
\displaystyle 2x_1+3y_1\le 30
\displaystyle 3x_1+y_1\le 17
\displaystyle x_1,\ y_1\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are } (0,10),\ (3,8) \text{ and } \!\left(\frac{17}{3},0\right).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=100x_1+120y_1\\ \hline (0,10) & 1200\\ \hline (3,8) & 1260\\ \hline \left(\frac{17}{3},0\right) & \frac{1700}{3}\\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 1260 \text{ which is attained at } (3,8).
\displaystyle \text{Thus, the maximum revenue is Rs } 1260 \text{ obtained when } 3 \text{ units of } x \text{ and  8 units of } \\ y \text{ were produced.}

Question 28.
A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of A and Rs 3 per unit of B. Each product is processed on two machines M₁ and M₂. One unit of A requires one minute of processing time on M₁ and two minutes of processing time on M₂, whereas one unit of B requires one minute of processing time on M₁ and one minute on M₂. Machines M₁ and M₂ are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce in a day in order to maximize the profit. Solve this problem graphically.  (CBSE 2000)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of product A and } y \text{ units of product B be manufactured.}
\displaystyle \text{Number of products cannot be negative.} \therefore\ x,\ y\ge 0
\displaystyle \text{According to the question, the given information can be tabulated as}
\displaystyle \begin{array}{|c|c|c|} \hline & \text{Time on } M_1\ (\text{minutes}) & \text{Time on } M_2\ (\text{minutes}) \\ \hline \text{Product A } (x) & 1 & 2 \\ \hline \text{Product B } (y) & 1 & 1 \\ \hline \text{Availability} & 300 & 360 \\ \hline \end{array}
\displaystyle \text{The constraints are}
\displaystyle x+y\le 300
\displaystyle 2x+y\le 360
\displaystyle \text{Firm manufactures two types of products A and B and sells them at a profit of Rs }5\text{ per unit} \\ \text{of type A and Rs }3\text{ per unit of type B.}
\displaystyle \text{Therefore, } x \text{ units of product A and } y \text{ units of product B give profit Rs }5x \\ \text{and Rs } 3y \text{ respectively.}
\displaystyle \text{Total profit } Z=5x+3y \text{ which is to be maximised.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Max } Z=5x+3y
\displaystyle \text{Subject to:}
\displaystyle x+y\le 300
\displaystyle 2x+y\le 360
\displaystyle x,\ y\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are } (0,0),\ (0,300),\ (60,240) \text{ and } (180,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=5x+3y \\ \hline (0,0) & 0 \\ \hline (0,300) & 900 \\ \hline (60,240)& 1020 \\ \hline (180,0) & 900 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is Rs } 1020 \text{ which is attained at } (60,240).
\displaystyle \text{Thus, the maximum profit is Rs } 1020 \text{ obtained when } 60 \text{ units of product A and } \\ 240 \text{units of product B are manufactured.}

Question 29.
A small firm manufactures items A and B. The total number of items A and B manufactured in a day is at most 24. Item A takes one hour to make while item B takes only half an hour. The maximum available time per day is 16 hours. If the profit on item A be Rs 300 and on item B be Rs 160, how many of each type of item should be produced to maximize the profit? Solve them graphically. (CBSE 2001, 2004)

\displaystyle \text{Answer:}
\displaystyle \text{Let the firm manufacture } x \text{ items of } A \text{ and } y \text{ items of } B \text{ per day.}
\displaystyle \text{Number of items cannot be negative.} \therefore\ x\ge 0 \text{ and } y\ge 0.
\displaystyle \text{It is given that the total number of items manufactured per day is at most } 24.
\displaystyle \therefore\ x+y\le 24.
\displaystyle \text{Item } A \text{ takes } 1 \text{ hour to make and item } B \text{ takes } 0.5 \text{ hour to make.}
\displaystyle \text{The maximum number of hours available per day is } 16 \text{ hours.}
\displaystyle \therefore\ x+0.5y\le 16.
\displaystyle \text{If the profit on one unit of item } A \text{ is Rs } 300 \text{ and one unit of item } B \text{ is Rs } 160.
\displaystyle \text{Therefore, profit gained on } x \text{ items of } A \text{ and } y \text{ items of } B \text{ is Rs } 300x \text{ and Rs } 160y \text{ respectively.}
\displaystyle \therefore\ \text{Total profit } Z=300x+160y.
\displaystyle \text{Thus, the mathematical formulation of the given problem is:}
\displaystyle \text{Maximise } Z=300x+160y
\displaystyle \text{Subject to:}
\displaystyle x+y\le 24
\displaystyle x+0.5y\le 16
\displaystyle x\ge 0,\ y\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are } (0,0),\ (16,0),\ (8,16) \text{ and } (0,24).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=300x+160y \\ \hline (0,0) & 0 \\ \hline (16,0) & 4800 \\ \hline (8,16) & 4960 \\ \hline (0,24) & 3840 \\ \hline \end{array}
\displaystyle \text{Thus, the maximum value of } Z \text{ is } 4960 \text{ at } (8,16).
\displaystyle \text{Thus, } 8 \text{ units of item } A \text{ and } 16 \text{ units of item } B \text{ should be manufactured per day to} \\ \text{maximise the profits.}

Question 30.
A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit? (CBSE 2001) 

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ toys of type } A \text{ and } y \text{ toys of type } B \text{ be manufactured.}
\displaystyle \text{The given information can be tabulated as follows:}
\displaystyle \begin{array}{|c|c|c|} \hline & \text{Cutting time (minutes)} & \text{Assembling time (minutes)} \\ \hline \text{Toy } A(x) & 5 & 10 \\ \hline \text{Toy } B(y) & 8 & 8 \\ \hline \text{Availability} & 180 & 240 \\ \hline \end{array}
\displaystyle \text{The constraints are}
\displaystyle 5x+8y\le 180
\displaystyle 10x+8y\le 240
\displaystyle \text{The profit is Rs }50\text{ each on type }A\text{ and Rs }60\text{ each on type }B.
\displaystyle \text{Therefore, profit gained on } x \text{ toys of type } A \text{ and } y \text{ toys of type } B \text{ is Rs } 50x  \\ \text{ and Rs } 60y \text{ respectively.}
\displaystyle \text{Total profit } Z=50x+60y
\displaystyle \text{The mathematical formulation of the given problem is}
\displaystyle \text{Max } Z=50x+60y \\ \text{Subject to:}
\displaystyle 5x+8y\le 180
\displaystyle 10x+8y\le 240
\displaystyle x\ge 0,\ y\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are } (0,0),\ \!\left(0,\frac{45}{2}\right),\ (12,15) \text{ and } (24,0).
\displaystyle \text{The values of } Z \text{ at the corner points are}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & Z=50x+60y \\ \hline (0,0) & 0 \\ \hline \left(0,\frac{45}{2}\right) & 1350 \\ \hline (12,15) & 1500 \\ \hline (24,0) & 1200 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 1500 \text{ which is attained at } (12,15).
\displaystyle \text{Thus, for maximum profit, } 12 \text{ units of toy } A \text{ and } 15 \text{ units of toy } B \text{ should be manufactured.}

Question 31.
A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments through which maximum capacity of the first department is 60 hours a week and that of the other department is 48 hours per week. The production of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit. (CBSE 2003)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units and } y \text{ units of articles } A \text{ and } B \text{ be produced respectively.}
\displaystyle \text{Number of articles cannot be negative. Hence, } \therefore\ x,\ y\ge 0
\displaystyle \text{The production of each unit of article } A \text{ requires } 4 \text{ hours in assembly and }\text{that of article }\\ B \text{ requires } 2 \text{ hours in assembly and the maximum capacity} \text{of the assembly department is } \\ 60 \text{ hours a week.}\Rightarrow 4x+2y\le 60
\displaystyle \text{The production of each unit of article } A \text{ requires } 2 \text{ hours in finishing and that of article } \\ B \text{ requires } 4 \text{ hours in finishing and the maximum capacity of the finishing department is } \\ 48 \text{ hours a week.}\Rightarrow 2x+4y\le 48
\displaystyle \text{If the profit is Rs } 6 \text{ for each unit of } A \text{ and Rs } 8 \text{ for each unit of } B.
\displaystyle \text{Therefore, profit gained from } x \text{ units and } y \text{ units of articles } A \text{ and } B \text{ respectively is Rs } \\ 6x \text{ and Rs } 8y \text{ respectively.}
\displaystyle \text{Total revenue } = Z=6x+8y \text{ which is to be maximised.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Max } Z=6x+8y
\displaystyle \text{Subject to:}
\displaystyle 2x+4y\le 48
\displaystyle 4x+2y\le 60
\displaystyle x,\ y\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are } (0,0),\ (0,12),\ (12,6) \text{ and } (15,0).
\displaystyle \text{The values of } Z \text{ at these corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=6x+8y \\ \hline (0,0) & 0 \\ \hline (0,12) & 96 \\ \hline (12,6) & 120 \\ \hline (15,0) & 90 \\ \hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 120 \text{ which is attained at } E_1(12,6).
\displaystyle \text{Thus, the maximum profit is Rs } 120 \text{ obtained when } 12 \text{ units of article } A \text{ and }\\ 6 \text{ units of article } B \text{ are manufactured.}

Question 32.
A firm makes A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and one and a half hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically. (CBSE 2004) 

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ and }y\text{ be the number of items of }A\text{ and }B\text{ that should be produced each day to} \\ \text{maximize the profit.}
\displaystyle \text{Number of items cannot be negative. Hence, }  x\ge0,\ y\ge0
\displaystyle \text{It is also given that the firm can produce at most }24\text{ items in a day. Hence, } \\ x+y\le24
\displaystyle \text{The time required to make an item of }A\text{ is one hour and time required to make and item} \\ \text{of }B\text{ is half an hour.}
\displaystyle \text{Therefore the time required to produce }x\text{ items of }A\text{ and }y\text{ items of }B \\ \text{is }x+\frac{1}{2}y\text{ hours.}
\displaystyle \text{However the maximum time available in a day is }16\text{ hours. Hence, } \\ x+\frac{1}{2}y\le16
\displaystyle \text{It is given that the profit on an item of }A\text{ is Rs }300\text{ and on one item of }B\text{ is Rs }160.
\displaystyle \text{Therefore the profit gained from }x\text{ items of }A\text{ and }y\text{ items of }B\text{ is Rs }300x\text{ and Rs }160y.
\displaystyle Z=300x+160y
\displaystyle \text{The mathematical form of the given LPP is:}
\displaystyle \text{Maximize }Z=300x+160y
\displaystyle \text{Subject to:}
\displaystyle x+y\le24
\displaystyle x+\frac{1}{2}y\le16
\displaystyle x\ge0,\ y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are }(0,0),\ (16,0),\ (8,16),\ (0,24)
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=300x+160y \\ \hline (0,0) & 0 \\ (16,0) & 4800 \\ (8,16) & 4960 \\ (0,24) & 3840 \\ \hline \end{array}
\displaystyle \text{Thus }x=8\text{ and }y=16\text{ and the maximum value of }Z\text{ is }4960

Question 33.
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and the maximum number of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for product A and Rs 15 per unit for product B, how many units of each product should be sold to maximize profit? 

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ units of product }A\text{ and }y\text{ units of product }B\text{ be manufactured. Hence, }  x\ge0,\ y\ge0
\displaystyle \text{It takes }5\text{ hours to produce one unit of }A\text{ and }3\text{ hours to produce one unit of }B.
\displaystyle \text{The two products are produced in a common production process with total capacity} \\ 500\text{ man-hours. Hence, }  5x+3y\le500
\displaystyle \text{The maximum number of units of }A\text{ that can be sold is }70\text{ and that of }B\text{ is }125.
\displaystyle x\le70
\displaystyle y\le125
\displaystyle \text{The profit per unit of }A\text{ is Rs }20\text{ and of }B\text{ is Rs }15.
\displaystyle \text{Hence profit from }x\text{ units of }A\text{ and }y\text{ units of }B\text{ is Rs }20x\text{ and Rs }15y.
\displaystyle Z=20x+15y
\displaystyle \text{The mathematical formulation of the given problem is}
\displaystyle \text{Max }Z=20x+15y
\displaystyle \text{Subject to:}
\displaystyle 5x+3y\le500
\displaystyle x\le70
\displaystyle y\le125
\displaystyle x\ge0,\ y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are }(0,0),(0,125),(25,125),(70,50),(70,0).
\displaystyle Z=20x+15y
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=20x+15y \\ \hline (0,0) & 0 \\ (0,125) & 1875 \\ (25,125) & 2375 \\ (70,50) & 2150 \\ (70,0) & 1400 \\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is }2375\text{ at } (25,125).
\displaystyle \text{Thus the maximum profit is Rs }2375\text{ when }25\text{ units of }A\text{ and }125\text{ units of }B\text{ are manufactured.}

Question 34.
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metres per box while the small boxes require 3 sq. metres per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock and if the profit on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ large boxes and }y\text{ small boxes be manufactured. Hence, } x\ge0,\;y\ge0
\displaystyle \text{The large boxes require }4\text{ sq. metre per box and the small boxes require } 3\text{ sq. metre} \\ \text{per box and }60\text{ sq. metre of cardboard is in stock. Hence, } 4x+3y\le60
\displaystyle \text{The manufacturer is required to make at least three large boxes and at least twice as} \\ \text{many small boxes as large boxes. Hence, }
\displaystyle x\ge3
\displaystyle y\ge2x
\displaystyle \text{If the profits on the large and small boxes are Rs }3\text{ and Rs }2\text{ per box respectively.}
\displaystyle Z=3x+2y
\displaystyle \text{Maximise }Z=3x+2y
\displaystyle \text{Subject to:}
\displaystyle 4x+3y\le60
\displaystyle x\ge3
\displaystyle y\ge2x
\displaystyle x\ge0,\;y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points are }(3,16),\;(6,12),\;(3,6).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=3x+2y \\ \hline (3,16) & 41 \\ (6,12) & 42 \\ C(3,6) & 21 \\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is }42\text{ which is at }(6,12).
\displaystyle \text{Thus for maximum profit is Rs }42,\;6\text{ large boxes and }12\text{ smaller boxes should be manufactured.}

Question 35.
A manufacturer makes two products, A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 units of product A and 16 units of product B. A working week consists of 40 hours of production and the weekly turnover must not be less than Rs 10000. If the profit on each product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum? Also find the maximum profit.

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ units of product }A\text{ and }y\text{ units of product }B\text{ be manufactured.}
\displaystyle \text{Number of units cannot be negative.} \text{Therefore, }x\ge 0,\ y\ge 0
\displaystyle \text{According to the question, the given information can be tabulated as:}
\displaystyle \begin{array}{|c|c|c|}\hline &\text{Selling price (Rs)}&\text{Manufacturing time (hrs)}\\ \hline \text{Product }A(x)&200&\ 1/2\\ \hline \text{Product }B(y)&300&1\\ \hline \end{array}
\displaystyle \text{Also, the availability of time is }40\text{ hours and the revenue should be at least Rs }10000.
\displaystyle \text{Further, it is given that there is a permanent order for }14\text{ units of product }A\text{ and} \\ 16\text{ units of product }B.
\displaystyle \text{Therefore, the constraints are}
\displaystyle 200x+300y\ge 10000
\displaystyle \frac{1}{2}x+y\le 40
\displaystyle x\ge 14
\displaystyle y\ge 16
\displaystyle \text{If the profit on each unit of product }A\text{ is Rs }20\text{ and on product }B\text{ is Rs }30.
\displaystyle \text{Therefore, profit gained on }x\text{ units of product }A\text{ and }y\text{ units of product } \\ B\text{ is Rs }20x\text{ and Rs }30y.
\displaystyle \text{Total profit }Z=20x+30y\text{ which is to be maximised.}
\displaystyle \text{Thus, the mathematical formulation of the given linear programming problem is}
\displaystyle \text{Maximise }Z=20x+30y
\displaystyle \text{Subject to:}
\displaystyle 2x+3y\ge 100
\displaystyle x+2y\le 80
\displaystyle x\ge 14
\displaystyle y\ge 16
\displaystyle x\ge 0,\ y\ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points of the feasible region are } (26,16),\  (48,16),\  (14,33)\text{ and } (14,24).
\displaystyle \text{The values of }Z\text{ at these corner points are}
\displaystyle \begin{array}{|c|c|}\hline \text{Corner point}&Z=20x+30y\\ \hline (26,16) &1000\\ \hline (48,16)&1440\\ \hline (14,33)&1270\\ \hline (14,24)&1000\\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is Rs }1440\text{ which is attained at } (48,16).
\displaystyle \text{Thus, the maximum profit is Rs }1440\text{ obtained when }48\text{ units of product }A\text{ and} \\ 16\text{ units of product }B\text{ were manufactured.}

Question 36.
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5 per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.  (CBSE 2007) 

\displaystyle \text{Answer:}
\displaystyle \text{Let us assume that the man travels } x\text{ km when the speed is } 25\text{ km/hour and } y\text{ km} \\ \text{when the speed is }40\text{ km/hour.}
\displaystyle \text{Thus, the total distance travelled is }(x+y)\text{ km.}
\displaystyle \text{Now, it is given that the man has Rs }100\text{ to spend on petrol. Hence, } 2x+5y\le100
\displaystyle \text{Now, time taken to travel }x\text{ km }=\frac{x}{25}\text{ h.}
\displaystyle \text{Time taken to travel }y\text{ km }=\frac{y}{40}\text{ h.}
\displaystyle \text{Now, it is given that the maximum time is }1\text{ hour. Hence, }
\displaystyle \frac{x}{25}+\frac{y}{40}\le1
\displaystyle \Rightarrow 8x+5y\le200
\displaystyle \text{Thus, the given linear programming problem is}
\displaystyle \text{Maximise }Z=x+y

\displaystyle \text{Subject to:}
\displaystyle 2x+5y\le100
\displaystyle 8x+5y\le200
\displaystyle x\ge0,\ y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The coordinates of the corner points of the feasible region are } (0,0),\ (25,0),\ \left(\frac{50}{3},\frac{40}{3}\right)\text{ and } (0,20).
\displaystyle \text{The value of the objective function at these points are given in the following table:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=x+y \\ \hline (0,0) & 0 \\ (25,0) & 25 \\ \left(\frac{50}{3},\frac{40}{3}\right) & 30 \\ (0,20) & 20 \\ \hline \end{array}
\displaystyle \text{So, the maximum value of }Z\text{ is }30\text{ at }x=\frac{50}{3},\ y=\frac{40}{3}
\displaystyle \text{Thus, the maximum distance that the man can travel in one hour is }30\text{ km.}
\displaystyle \text{Hence, the distance travelled at }25\text{ km/hour is }\frac{50}{3}\text{ km and at }40\text{ km/hour is }\frac{40}{3}\text{ km.}

Question 37.
An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:

\displaystyle \begin{array}{|c|c|c|} \hline \text{To} & A & B \\ \hline D & 7 & 3 \\ E & 6 & 4 \\ F & 3 & 2 \\ \hline \end{array}

Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum? 

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ and }y\text{ litres of oil be supplied from }A\text{ to the petrol pumps }D\text{ and }E.
\displaystyle \text{Then }(7000-x-y)\text{ litres will be supplied from }A\text{ to petrol pump }F.
\displaystyle \text{The requirement at petrol pump }D\text{ is }4500\text{ litres.}
\displaystyle \text{Since }x\text{ litres are transported from depot }A,\text{ the remaining }(4500-x) \text{ litres will be} \\ \text{transported from petrol pump }B.
\displaystyle \text{Similarly }(3000-y)\text{ litres and }\left[3500-(7000-x-y)\right]\text{ litres i.e. }(x+y-3500)\text{ litres} \\ \text{will be transported from depot }B\text{ to petrol pumps }E\text{ and }F\text{ respectively.}
\displaystyle \text{The given problem can be represented diagrammatically as follows.}
\displaystyle x\ge0,\ y\ge0,\ \text{and }(7000-x-y)\ge0
\displaystyle \Rightarrow x\ge0,\ y\ge0,\ \text{and }x+y\le7000
\displaystyle 4500-x\ge0,\ 3000-y\ge0,\ \text{and }x+y-3500\ge0
\displaystyle \Rightarrow x\le4500,\ y\le3000,\ \text{and }x+y\ge3500
\displaystyle \text{Cost of transporting }10\text{ L of petrol }=\text{Re }1
\displaystyle \text{Cost of transporting }1\text{ L of petrol }=\text{Rs }\frac{1}{10}
\displaystyle \text{Therefore, total transportation cost is given by}
\displaystyle z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y) \\ =\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)
\displaystyle =0.3x+0.1y+3950
\displaystyle \text{The problem can be formulated as follows.}
\displaystyle \text{Minimize }Z=0.3x+0.1y+3950
\displaystyle \text{Subject to:}
\displaystyle x+y\le7000
\displaystyle x\le4500
\displaystyle y\le3000
\displaystyle x+y\ge3500
\displaystyle x\ge0,\ y\ge0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points are }(3500,0),\ (4500,0),\ (4500,2500),\ (4000,3000),\ (500,3000)
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=0.3x+0.1y+3950 \\ \hline (3500,0) & 5000 \\ (4500,0) & 5300 \\ (4500,2500) & 5550 \\ (4000,3000) & 5450 \\ (500,3000) & 4400 \\ \hline \end{array}
\displaystyle \text{The minimum value of }Z\text{ is }4400\text{ at } (500,3000)
\displaystyle \text{The minimum transportation cost is Rs }4400

Question 38.
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day so as to earn the maximum profit. Make it as an LPP and solve it graphically. (CBSE 2010) 

\displaystyle \text{Answer:}
\displaystyle \text{Let the firm manufacture } x \text{ gold rings and } y \text{ chains per day.}
\displaystyle \text{Since the numbers cannot be negative,}  x \ge 0,\ y \ge 0
\displaystyle \text{The total number produced per day is at most } 24. \Rightarrow  x+y \le 24
\displaystyle \text{A ring takes } 1 \text{ hour and a chain takes } 0.5 \text{ hour to make.}
\displaystyle \text{The maximum number of working hours per day is } 16. \Rightarrow  x+0.5y \le 16
\displaystyle \text{Profit on a ring is Rs }300 \text{ and on a chain is Rs }190.
\displaystyle Z = 300x + 190y
\displaystyle \text{Maximise } Z = 300x + 190y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 24
\displaystyle x+0.5y \le 16
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (16,0),\ (8,16),\ (0,24).
\displaystyle Z(0,0)=300(0)+190(0)=0
\displaystyle Z(16,0)=300(16)+190(0)=4800
\displaystyle Z(8,16)=300(8)+190(16)=5440
\displaystyle Z(0,24)=300(0)+190(24)=4560
\displaystyle \text{The maximum value of } Z \text{ is } 5440 \text{ at } (8,16).
\displaystyle \text{Hence, } 8 \text{ gold rings and } 16 \text{ chains should be produced per day.}

Question 39.
A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weigh 1 kg and 1½ kg each respectively. The shelf is 96 cm long and at most can support a weight of 21 kg. How should the shelf be filled with the books of two types in order to include the greatest number of books? Make it as an LPP and solve it graphically. (CBSE 2010)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ books of first type and } y \text{ books of second type be accommodated.}
\displaystyle \text{Since the number of books cannot be negative, } \ x \ge 0,\ y \ge 0
\displaystyle \text{According to the question, the given information is tabulated below:}
\displaystyle \text{Thickness (cm) and Weight (kg) for first type: } 6,\ 1
\displaystyle \text{Thickness (cm) and Weight (kg) for second type: } 4,\ 1.5
\displaystyle \text{Capacity of shelf for thickness and weight: } 96,\ 21
\displaystyle \text{Therefore, the constraints are}
\displaystyle 6x+4y \le 96
\displaystyle x+1.5y \le 21
\displaystyle \text{Number of books } = Z = x+y \text{ which is to be maximised}
\displaystyle \text{Maximise } Z = x+y
\displaystyle \text{Subject to:}
\displaystyle 6x+4y \le 96
\displaystyle x+1.5y \le 21
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,14),\ (12,6),\ (16,0).
\displaystyle Z(0,0)=0+0=0
\displaystyle Z(0,14)=0+14=14
\displaystyle Z(12,6)=12+6=18
\displaystyle Z(16,0)=16+0=16
\displaystyle \text{The maximum value of } Z \text{ is } 18 \text{ at } (12,6).
\displaystyle \text{Hence, the maximum number of books that can be arranged is } 18.
\displaystyle \text{Therefore, } 12 \text{ books of the first type and } 6 \text{ books of the second type should be arranged.}

Question 40.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the number of tennis rackets and cricket bats the factory must manufacture to earn the maximum profit. Make it as an LPP and solve it graphically. (CBSE 2011) 

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ be the number of tennis rackets and } y \text{ be the number of cricket bats sold.}
\displaystyle \text{Since numbers cannot be negative. Hence, } x \ge 0,\ y \ge 0
\displaystyle \text{A tennis racket requires } 1.5 \text{ hours of machine time and } 3 \text{ hours of craftsman's time,}
\displaystyle \text{while a cricket bat requires } 3 \text{ hours of machine time and } 1 \text{ hour of craftsman's time.}
\displaystyle \text{The factory has at most } 42 \text{ hours of machine time and } 24 \text{ hours of craftsman's time available.}
\displaystyle 1.5x+3y \le 42
\displaystyle 3x+y \le 24
\displaystyle \text{Profit on a racket is Rs }20 \text{ and on a bat is Rs }10.
\displaystyle Z = 20x + 10y
\displaystyle \text{Maximise } Z = 20x + 10y
\displaystyle \text{Subject to:}
\displaystyle 1.5x+3y \le 42
\displaystyle 3x+y \le 24
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,14),\ (4,12),\ (8,0).
\displaystyle Z(0,0)=20(0)+10(0)=0
\displaystyle Z(0,14)=20(0)+10(14)=140
\displaystyle Z(4,12)=20(4)+10(12)=200
\displaystyle Z(8,0)=20(8)+10(0)=160
\displaystyle \text{The maximum value of } Z \text{ is } 200 \text{ at } (4,12).
\displaystyle \text{Hence, the maximum profit is Rs }200.
\displaystyle \text{Therefore, } 4 \text{ tennis rackets and } 12 \text{ cricket bats should be sold.}

Question 41.
A merchant plans to sell two types of personal computers: a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakh and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make it as an LPP and solve it graphically. (CBSE 2011)

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ and } y \text{ be the number of desktop models and portable models respectively.}
\displaystyle \text{Since numbers cannot be negative, Hence } x \ge 0,\ y \ge 0
\displaystyle \text{The monthly demand will not exceed } 250 \text{ units. Hence, } x+y \le 250
\displaystyle \text{Cost of a desktop model is Rs }25000 \text{ and that of a portable model is Rs }40000.
\displaystyle \text{The total investment cannot exceed Rs }7000000.
\displaystyle 25000x+40000y \le 7000000
\displaystyle \text{Profit on a desktop model is Rs }4500 \text{ and on a portable model is Rs }5000.
\displaystyle Z = 4500x + 5000y
\displaystyle \text{Maximise } Z = 4500x + 5000y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 250
\displaystyle 25000x+40000y \le 7000000
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,175),\ (200,50),\ (250,0).
\displaystyle Z(0,0)=4500(0)+5000(0)=0
\displaystyle Z(0,175)=4500(0)+5000(175)=875000
\displaystyle Z(200,50)=4500(200)+5000(50)=1150000
\displaystyle Z(250,0)=4500(250)+5000(0)=1125000
\displaystyle \text{The maximum value of } Z \text{ is } 1150000 \text{ at } (200,50).
\displaystyle \text{Hence, the maximum profit is Rs }1150000.
\displaystyle \text{Therefore, } 200 \text{ desktop models and } 50 \text{ portable models should be sold.}

Question 42.
A cooperative society of farmers has 50 hectare land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society? (CBSE 2013)

\displaystyle \text{Answer:}
\displaystyle \text{Let the land allocated for crop X be } x \text{ hectares and for crop Y be } y \text{ hectares.}
\displaystyle \text{The maximum area of land available for two crops is } 50 \text{ hectares. Hence, }
\displaystyle x+y \le 50
\displaystyle \text{The liquid herbicide required is } 20 \text{ litres per hectare for crop X and } 10 \text{ litres per hectare} \\ \text{for crop Y.}
\displaystyle \text{The maximum amount of herbicide available is } 800 \text{ litres. Hence, }
\displaystyle 20x+10y \le 800 \Rightarrow  2x+y \le 80
\displaystyle \text{The profit from crop X is Rs }10500 \text{ per hectare and from crop Y is Rs }9000 \text{ per hectare.}
\displaystyle Z = 10500x + 9000y
\displaystyle \text{Maximise } Z = 10500x + 9000y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 50
\displaystyle 2x+y \le 80
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points of the feasible region are } (0,0),\ (40,0),\ (30,20),\ (0,50).
\displaystyle Z(0,0)=10500(0)+9000(0)=0
\displaystyle Z(40,0)=10500(40)+9000(0)=420000
\displaystyle Z(30,20)=10500(30)+9000(20)=495000
\displaystyle Z(0,50)=10500(0)+9000(50)=450000
\displaystyle \text{The maximum value of } Z \text{ is } 495000 \text{ at } (30,20).
\displaystyle \text{Hence, } 30 \text{ hectares should be allocated to crop X and } 20 \text{ hectares to crop Y.}
\displaystyle \text{Therefore, the maximum profit of the society is Rs }495000.

Question 43.
A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 8000 on each piece of Model A and Rs 12000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise maximum profit? What is the maximum profit per week?

\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ pieces of model A and } y \text{ pieces of model B are manufactured per week.}
\displaystyle \text{Each piece of model A requires } 9 \text{ labour hours and each piece of model B requires} \\ 12 \text{ labour hours for fabricating.}
\displaystyle \text{Hence, the total labour hours required for fabrication are } 9x+12y.
\displaystyle \text{The maximum labour hours available for fabrication are } 180. \text{Hence, }
\displaystyle 9x+12y \le 180\Rightarrow 3x+4y \le 60
\displaystyle \text{Each piece of model A requires } 1 \text{ labour hour and each piece of model B requires} \\ 3 \text{ labour hours for finishing.}
\displaystyle \text{Hence, the total labour hours required for finishing are } x+3y.
\displaystyle \text{The maximum labour hours available for finishing are } 30.
\displaystyle x+3y \le 30
\displaystyle \text{Profit from each piece of model A is Rs }8000 \text{ and from each piece of model B is Rs }12000.
\displaystyle Z = 8000x + 12000y
\displaystyle \text{Maximise } Z = 8000x + 12000y
\displaystyle \text{Subject to:}
\displaystyle 3x+4y \le 60
\displaystyle x+3y \le 30
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (20,0),\ (12,6),\ (0,10).
\displaystyle Z(0,0)=8000(0)+12000(0)=0
\displaystyle Z(20,0)=8000(20)+12000(0)=160000
\displaystyle Z(12,6)=8000(12)+12000(6)=168000
\displaystyle Z(0,10)=8000(0)+12000(10)=120000
\displaystyle \text{The maximum value of } Z \text{ is } 168000 \text{ at } (12,6).
\displaystyle \text{Hence, the manufacturing company should produce } 12 \text{ pieces of model A and} \\ 6 \text{ pieces of model B.}
\displaystyle \text{Therefore, the maximum profit is Rs }168000.

Question 44.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ be the number of tennis rackets and } y \text{ be the number of cricket bats sold.}
\displaystyle \text{Since numbers cannot be negative,}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{A tennis racket requires } 1.5 \text{ hours of machine time and } 3 \text{ hours of craftsman's time,}
\displaystyle \text{while a cricket bat requires } 3 \text{ hours of machine time and } 1 \text{ hour of craftsman's time.}
\displaystyle \text{The factory has at most } 42 \text{ hours of machine time and } 24 \text{ hours of craftsman's time available.}
\displaystyle 1.5x+3y \le 42
\displaystyle 3x+y \le 24
\displaystyle \text{Profit on a racket is Rs }20 \text{ and on a bat is Rs }10.
\displaystyle Z = 20x + 10y
\displaystyle \text{Maximise } Z = 20x + 10y
\displaystyle \text{Subject to:}
\displaystyle 1.5x+3y \le 42
\displaystyle 3x+y \le 24
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,14),\ (4,12),\ (8,0).
\displaystyle Z(0,0)=20(0)+10(0)=0
\displaystyle Z(0,14)=20(0)+10(14)=140
\displaystyle Z(4,12)=20(4)+10(12)=200
\displaystyle Z(8,0)=20(8)+10(0)=160
\displaystyle \text{The maximum value of } Z \text{ is } 200 \text{ at } (4,12).
\displaystyle \text{Hence, } 4 \text{ tennis rackets and } 12 \text{ cricket bats must be made for full capacity.}
\displaystyle \text{Therefore, the maximum profit of the factory is Rs }200.

Question 45.
A merchant plans to sell two types of personal computers: a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ and } y \text{ be the number of desktop models and portable models respectively.}
\displaystyle \text{Since numbers cannot be negative, } x \ge 0,\ y \ge 0
\displaystyle \text{The monthly demand will not exceed } 250 \text{ units. Hence, } x+y \le 250
\displaystyle \text{Cost of a desktop model is Rs }25000 \text{ and that of a portable model is Rs }40000.
\displaystyle \text{The total investment cannot exceed Rs }7000000. \\ \text{ Hence,  }  25000x+40000y \le 7000000
\displaystyle \text{Profit on a desktop model is Rs }4500 \text{ and on a portable model is Rs }5000.
\displaystyle Z = 4500x + 5000y
\displaystyle \text{Maximise } Z = 4500x + 5000y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 250
\displaystyle 25000x+40000y \le 7000000
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,175),\ (200,50),\ (250,0).
\displaystyle Z(0,0)=4500(0)+5000(0)=0
\displaystyle Z(0,175)=4500(0)+5000(175)=875000
\displaystyle Z(200,50)=4500(200)+5000(50)=1150000
\displaystyle Z(250,0)=4500(250)+5000(0)=1125000
\displaystyle \text{The maximum value of } Z \text{ is } 1150000 \text{ at } (200,50).
\displaystyle \text{Hence, } 200 \text{ desktop models and } 50 \text{ portable models should be sold.}

Question 46.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 per doll and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of doll A and } y \text{ units of doll B be manufactured to obtain the maximum profit.}
\displaystyle \text{The mathematical formulation of the problem is as follows:}
\displaystyle \text{Maximise } Z = 12x + 16y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 1200
\displaystyle y \le \frac{x}{2}
\displaystyle x-3y \le 600
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (800,400),\ (1050,150),\ (600,0).
\displaystyle Z(0,0)=12(0)+16(0)=0
\displaystyle Z(800,400)=12(800)+16(400)=16000
\displaystyle Z(1050,150)=12(1050)+16(150)=15000
\displaystyle Z(600,0)=12(600)+16(0)=7200
\displaystyle \text{The maximum value of } Z \text{ is } 16000 \text{ at } (800,400).
\displaystyle \text{Hence, } 800 \text{ units of doll A and } 400 \text{ units of doll B should be produced weekly to obtain} \\ \text{the maximum profit of Rs }16000.

Question 47.
There are two types of fertilisers F₁ and F₂. F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If F₁ costs Rs 6 per kg and F₂ costs Rs 5 per kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at minimum cost. What is the minimum cost?

\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ kg of fertilizer } F_1 \text{ and } y \text{ kg of fertilizer } F_2 \text{ are used to meet nutrient requirements.}
\displaystyle \text{F}_1 \text{ contains } 10\% \text{ nitrogen and } \text{F}_2 \text{ contains } 5\% \text{ nitrogen.}
\displaystyle \text{The farmer needs at least } 14 \text{ kg of nitrogen. Hence, } \\ \frac{x}{10}+\frac{y}{20} \ge 14\Rightarrow 2x+y \ge 280
\displaystyle \text{Similarly, } \text{F}_1 \text{ contains } 6\% \text{ phosphoric acid and } \text{F}_2 \text{ contains } 10\% \text{ phosphoric acid.}
\displaystyle \text{The farmer needs at least } 14 \text{ kg of phosphoric acid. Hence, }
\displaystyle \frac{6x}{100}+\frac{10y}{100} \ge 14 \Rightarrow 3x+5y \ge 700
\displaystyle \text{The cost of fertilizer } F_1 \text{ is Rs }6 \text{ per kg and that of } F_2 \text{ is Rs }5 \text{ per kg.}
\displaystyle Z = 6x+5y
\displaystyle \text{Minimise } Z = 6x+5y
\displaystyle \text{Subject to:}
\displaystyle 2x+y \ge 280
\displaystyle 3x+5y \ge 700
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } \!\left(\frac{700}{3},0\right),\ (100,80),\ (0,280).
\displaystyle Z\!\left(\frac{700}{3},0\right)=6\!\left(\frac{700}{3}\right)+5(0)=1400
\displaystyle Z(100,80)=6(100)+5(80)=1000
\displaystyle Z(0,280)=6(0)+5(280)=1400
\displaystyle \text{The minimum value of } Z \text{ is } 1000 \text{ at } (100,80).
\displaystyle \text{Hence, } 100 \text{ kg of fertilizer } F_1 \text{ and } 80 \text{ kg of fertilizer } F_2 \text{ should be used.}
\displaystyle \text{Therefore, the minimum cost is Rs }1000.

Question 48.
A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for at least 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Item} & \text{Machine I} & \text{Machine II} & \text{Machine III} \\ \hline M & 1 & 2 & 1 \\ \hline N & 2 & 1 & \frac{5}{4} \\ \hline \end{array}

She makes a profit of Rs 600 and Rs 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?

\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ units of item M and } y \text{ units of item N are produced to maximise the profit.}
\displaystyle \text{Each unit of item M requires } 1 \text{ hour on machine I and each unit of item N requires} \\ 2 \text{vhours on machine I.}
\displaystyle \text{Hence, the total hours required on machine I are } 2x+y.
\displaystyle \text{Machine I can be operated for at most } 12 \text{ hours. Hence, } 2x+y \le 12
\displaystyle \text{Each unit of item M requires } 2 \text{ hours on machine II and each unit of item N requires} \\ 1 \text{vhour on machine II.}
\displaystyle \text{Hence, the total hours required on machine II are } x+2y.
\displaystyle \text{Machine II can be operated for at most } 12 \text{ hours. Hence, } x+2y \le 12
\displaystyle \text{Each unit of item M requires } 1 \text{ hour on machine III and each unit of item N requires} \\ 1.25 \text{ hours on machine III.}
\displaystyle \text{Hence, the total hours required on machine III are } x+1.25y.
\displaystyle \text{Machine III must be operated for at least } 5 \text{ hours. Hence, } x+1.25y \ge 5
\displaystyle \text{The profit from each unit of item M is Rs }600 \text{ and from each unit of item N is Rs }400.
\displaystyle Z = 600x + 400y
\displaystyle \text{Maximise } Z = 600x + 400y
\displaystyle \text{Subject to:}
\displaystyle 2x+y \le 12
\displaystyle x+2y \le 12
\displaystyle x+1.25y \ge 5
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}

\displaystyle \text{The corner points of the feasible region are } (5,0),\ (6,0),\ (4,4),\ (0,6),\ (0,4).
\displaystyle Z(5,0)=600(5)+400(0)=3000
\displaystyle Z(6,0)=600(6)+400(0)=3600
\displaystyle Z(4,4)=600(4)+400(4)=4000
\displaystyle Z(0,6)=600(0)+400(6)=2400
\displaystyle Z(0,4)=600(0)+400(4)=1600
\displaystyle \text{The maximum value of } Z \text{ is } 4000 \text{ at } (4,4).
\displaystyle \text{Hence, } 4 \text{ units of item M and } 4 \text{ units of item N should be produced.}
\displaystyle \text{Therefore, the maximum profit of the manufacturer is Rs }4000.

Question 49.
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacities of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{From} & A & B & C \\ \hline P & 160 & 100 & 150 \\ Q & 100 & 120 & 100 \\ \hline \end{array}

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? What will be the minimum transportation cost?

\displaystyle \text{Answer:}
\displaystyle \text{Here, total demand of the commodity is } (5+5+4)=14 \text{ units and total supply is} \\ (8+6)=14 \text{ units, so no commodity is left at the factories.}
\displaystyle \text{Let } x \text{ and } y \text{ units of the commodity be transported from factory } \text{P to depots A and} \\ \text{B respectively.}
\displaystyle \text{Then } 8-x-y \text{ units are transported from factory P to depot C.}
\displaystyle \text{The weekly requirement of depot A is } 5 \text{ units, so } 5-x \text{ units are transported from factory} \\ \text{Q to depot A.}
\displaystyle \text{The weekly requirement of depot B is } 5 \text{ units, so } 5-y \text{ units are transported from factory} \\ \text{Q to depot B.}
\displaystyle \text{The weekly requirement of depot C is } 4 \text{ units, hence,} \\ 6-(5-x)-(5-y)=x+y-4 \text{ units are transported from factory Q to depot C.}
\displaystyle \text{Since transported quantities cannot be negative,}
\displaystyle x \ge 0,\ y \ge 0,\ 8-x-y \ge 0,\ 5-x \ge 0,\ 5-y \ge 0,\ x+y-4 \ge 0
\displaystyle x \ge 0,\ y \ge 0,\ x+y \le 8,\ x \le 5,\ y \le 5,\ x+y \ge 4
\displaystyle \text{Total transportation cost is}
\displaystyle 160x+100y+150(8-x-y)+100(5-x)+120(5-y)+100(x+y-4)
\displaystyle Z = 10x-70y+1900
\displaystyle \text{Minimise } Z = 10x-70y+1900
\displaystyle \text{Subject to:}
\displaystyle x+y \le 8
\displaystyle x \le 5
\displaystyle y \le 5
\displaystyle x+y \ge 4
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (4,0),\ (5,0),\ (5,3),\ (3,5),\ (0,5),\ (0,4).
\displaystyle \text{The values of the objective function at these points are:}
\displaystyle Z(4,0)=10(4)-70(0)+1900=1940
\displaystyle Z(5,0)=10(5)-70(0)+1900=1950
\displaystyle Z(5,3)=10(5)-70(3)+1900=1740
\displaystyle Z(3,5)=10(3)-70(5)+1900=1580
\displaystyle Z(0,5)=10(0)-70(5)+1900=1550
\displaystyle Z(0,4)=10(0)-70(4)+1900=1620
\displaystyle \text{The minimum value of } Z \text{ is } 1550 \text{ at } (0,5).
\displaystyle \text{Hence, for minimum transportation cost, factory P supplies } 0,\ 5,\ 3 \text{ units to depots A, B} \\ \text{and C respectively.}
\displaystyle \text{Factory Q supplies } 5,\ 0,\ 1 \text{ units to depots A, B and C respectively.}
\displaystyle \text{Therefore, the minimum transportation cost is Rs }1550.

Question 50.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time in minutes required for each toy on the machines is given below:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Type} & I & II & III \\ \hline A & 12 & 18 & 6 \\ B & 6 & 0 & 9 \\ \hline \end{array}

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

\displaystyle \text{Answer:}
\displaystyle \text{Suppose the manufacturer makes } x \text{ toys of type A and } y \text{ toys of type B.}
\displaystyle \text{Each toy of type A requires } 12 \text{ minutes on machine I and each toy of type B requires} \\ 6 \text{ minutes on machine I.}
\displaystyle \text{Hence, the total time required on machine I is } 12x+6y \text{ minutes.}
\displaystyle \text{Machine I is available for at most } 360 \text{ minutes. Hence, } 12x+6y \le 360 \Rightarrow 2x+y \le 60
\displaystyle \text{Each toy of type A requires } 18 \text{ minutes on machine II and each toy of type B requires } \\  0 \text{ minutes on machine II.}
\displaystyle \text{Hence, the total time required on machine II is } 18x.
\displaystyle \text{Machine II is available for at most } 360 \text{ minutes. Hence, } 18x \le 360\Rightarrow x \le 20
\displaystyle \text{Each toy of type A requires } 6 \text{ minutes on machine III and each toy of type B requires } \\  9 \text{ minutes on machine III.}
\displaystyle \text{Hence, the total time required on machine III is } 6x+9y \text{ minutes.}
\displaystyle \text{Machine III is available for at most } 360 \text{ minutes. Hence, } 6x+9y \le 360 \Rightarrow 2x+3y \le 120
\displaystyle \text{The profit on each toy of type A is Rs }7.50 \text{ and on each toy of type B is Rs }5.
\displaystyle Z = 7.5x + 5y
\displaystyle \text{Maximise } Z = 7.5x + 5y
\displaystyle \text{Subject to:}
\displaystyle 2x+y \le 60
\displaystyle x \le 20
\displaystyle 2x+3y \le 120
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}
\displaystyle \text{The corner points of the feasible region are } (0,0),\ (20,0),\ (20,20),\ (15,30),\ (0,40).
\displaystyle Z(0,0)=7.5(0)+5(0)=0
\displaystyle Z(20,0)=7.5(20)+5(0)=150
\displaystyle Z(20,20)=7.5(20)+5(20)=250
\displaystyle Z(15,30)=7.5(15)+5(30)=262.5
\displaystyle Z(0,40)=7.5(0)+5(40)=200
\displaystyle \text{The maximum value of } Z \text{ is } 262.5 \text{ at } (15,30).
\displaystyle \text{Hence, } 15 \text{ toys of type A and } 30 \text{ toys of type B should be manufactured per day.}
\displaystyle \text{Therefore, the maximum profit is Rs }262.50.

Question 51.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by executive. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?

\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ tickets of executive class and } y \text{ tickets of economy class are sold by the airline.}
\displaystyle \text{The profit on each executive class ticket is Rs }1000 \text{ and on each economy class ticket is Rs }600.
\displaystyle \text{Hence, the total profit is} \ Z = 1000x + 600y
\displaystyle \text{The aeroplane can carry at most } 200 \text{ passengers. Hence, } x+y \le 200
\displaystyle \text{The airline reserves at least } 20 \text{ seats for executive class. Hence, } x \ge 20
\displaystyle \text{At least four times as many passengers prefer economy class as executive class.}
\displaystyle y \ge 4x
\displaystyle \text{Maximise } Z = 1000x + 600y
\displaystyle \text{Subject to:}
\displaystyle x+y \le 200
\displaystyle x \ge 20
\displaystyle y \ge 4x
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (20,80),\ (40,160),\ (20,180).
\displaystyle Z(20,80)=1000(20)+600(80)=68000
\displaystyle Z(40,160)=1000(40)+600(160)=136000
\displaystyle Z(20,180)=1000(20)+600(180)=128000
\displaystyle \text{The maximum value of } Z \text{ is } 136000 \text{ at } (40,160).
\displaystyle \text{Hence, } 40 \text{ executive class tickets and } 160 \text{ economy class tickets should be sold.}
\displaystyle \text{Therefore, the maximum profit of the airline is Rs }136000.

Question 52.
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at Rs 100 and Rs 120 per unit respectively, how should he use his resources to maximize total revenue? Formulate the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and should be paid at the same rate?

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of A and } y \text{ units of B be produced by the manufacturer.}
\displaystyle \text{The price of one unit of A is Rs }100 \text{ and the price of one unit of B is Rs }120.
\displaystyle \text{Hence, the total revenue is:}   \displaystyle Z = 100x + 120y
\displaystyle \text{One unit of A requires } 2 \text{ workers and one unit of B requires } 3 \text{ workers.}
\displaystyle \text{Hence, total workers required are } 2x+3y.
\displaystyle \text{The manufacturer has } 30 \text{ workers available. Hence,  } 2x+3y \le 30
\displaystyle \text{One unit of A requires } 3 \text{ units of capital and one unit of B requires } 1 \text{ unit of capital.}
\displaystyle \text{Hence, total capital required is } 3x+y.
\displaystyle \text{The manufacturer has } 17 \text{ units of capital available. Hence,  } 3x+y \le 17
\displaystyle \text{Maximise } Z = 100x + 120y
\displaystyle \text{Subject to:}
\displaystyle 2x+3y \le 30
\displaystyle 3x+y \le 17
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (0,10),\ \!\left(\frac{17}{3},0\right),\ (3,8).
\displaystyle Z(0,0)=100(0)+120(0)=0
\displaystyle Z(0,10)=100(0)+120(10)=1200
\displaystyle Z\!\left(\frac{17}{3},0\right)=100\!\left(\frac{17}{3}\right)+120(0)=\frac{1700}{3}
\displaystyle Z(3,8)=100(3)+120(8)=1260
\displaystyle \text{The maximum value of } Z \text{ is } 1260 \text{ at } (3,8).
\displaystyle \text{Hence, the maximum total revenue is Rs }1260 \text{ when } 3 \text{ units of A and } 8 \text{ units of B are} \\ \text{produced.}
\displaystyle \text{Yes, because the efficiency of a worker does not depend on whether the worker is} \\ \text{male or female.}

Question 53.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs 7 profit and that of B at a profit of Rs 4. Find the production level per day for maximum profit graphically.

\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ units of product A and } y \text{ units of product B be manufactured per day.}
\displaystyle \text{One unit of product A requires } 3 \text{ hours on the first machine and one unit of product} \\ \text{B requires } 2 \text{ hours on the first machine.}
\displaystyle \text{The first machine is available for } 12 \text{ hours per day.}     \displaystyle 3x+2y \le 12
\displaystyle \text{One unit of product A requires } 3 \text{ hours on the second machine and one unit of product} \\ \text{B requires } 1 \text{ hour on the second machine.}
\displaystyle \text{The second machine is available for } 9 \text{ hours per day.}     \displaystyle 3x+y \le 9
\displaystyle \text{The profit on product A is Rs }7 \text{ per unit and on product B is Rs }4 \text{ per unit.}
\displaystyle Z = 7x + 4y
\displaystyle \text{Maximise } Z = 7x + 4y
\displaystyle \text{Subject to:}
\displaystyle 3x+2y \le 12
\displaystyle 3x+y \le 9
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The corner points of the feasible region are } (0,0),\ (3,0),\ (2,3),\ (0,6).
\displaystyle Z(0,0)=7(0)+4(0)=0
\displaystyle Z(3,0)=7(3)+4(0)=21
\displaystyle Z(2,3)=7(2)+4(3)=26
\displaystyle Z(0,6)=7(0)+4(6)=24
\displaystyle \text{The maximum value of } Z \text{ is } 26 \text{ at } (2,3).
\displaystyle \text{Hence, } 2 \text{ units of product A and } 3 \text{ units of product B should be manufactured per day.}
\displaystyle \text{Therefore, the maximum daily profit of the manufacturer is Rs }26.

Question 54.
There are two types of fertilisers A′ and B′. A′ consists of 12% nitrogen and 5% phosphoric acid while B′ consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, a farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crop. If A′ costs Rs 10 per kg and B′ costs Rs 8 per kg, determine graphically how much of each type of fertiliser should be used so that nutrient requirements are met at minimum cost.

\displaystyle \text{Answer:}
\displaystyle \text{The given information can be tabulated as follows:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Fertilizer} & \text{Nitrogen} & \text{Phosphoric Acid} & \text{Cost/kg (in Rs)}\\\hline A & 12\% & 5\% & 10\\ B & 4\% & 5\% & 8\\\hline \end{array}
\displaystyle \text{Let the requirement of fertilizer A be } x \text{ kg and that of fertilizer B be } y \text{ kg.}
\displaystyle \text{The farmer requires at least } 12 \text{ kg of nitrogen and } 12 \text{ kg of phosphoric acid.}
\displaystyle \frac{12}{100}x+\frac{4}{100}y \ge 12
\displaystyle \Rightarrow 12x+4y \ge 1200
\displaystyle \Rightarrow 3x+y \ge 300
\displaystyle \text{Also,  }   \frac{5}{100}x+\frac{5}{100}y \ge 12
\displaystyle \Rightarrow 5x+5y \ge 1200
\displaystyle \Rightarrow  x+y \ge 240
\displaystyle \text{The total cost of the fertilizers is}
\displaystyle Z = 10x + 8y
\displaystyle \text{Minimise } Z = 10x + 8y
\displaystyle \text{Subject to:}
\displaystyle 3x+y \ge 300
\displaystyle x+y \ge 240
\displaystyle x \ge 0,\ y \ge 0

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle \text{The feasible region is unbounded.} \\ \text{The corner points considered are } (0,300),\ (30,210),\ (240,0).
\displaystyle Z(0,300)=10(0)+8(300)=2400
\displaystyle Z(30,210)=10(30)+8(210)=1980
\displaystyle Z(240,0)=10(240)+8(0)=2400
\displaystyle \text{The minimum value of } Z \text{ is } 1980 \text{ at } (30,210).
\displaystyle \text{Hence, the minimum requirement of fertilizer A is } 30 \text{ kg and that of fertilizer B is } 210 \text{ kg.}
\displaystyle \text{Therefore, the minimum total cost of the fertilizers is Rs }1980.

Question 55.
A small firm manufactures necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300, formulate an LPP for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced daily.

\displaystyle \text{Answer:}
\displaystyle x=\text{ number of necklaces per day}
\displaystyle y=\text{ number of bracelets per day}
\displaystyle \text{Maximize } Z=100x+300y
\displaystyle \text{Subject to:}
\displaystyle x+y\le 24
\displaystyle 0.5x+y\le16
\displaystyle x+2y\le32
\displaystyle x\ge1
\displaystyle y\ge1
\displaystyle \text{Corner points of feasible region:}
\displaystyle (x+2y=32)-(x+y=24)\Rightarrow y=8
\displaystyle x=16 \Rightarrow (16,8)
\displaystyle x=1,\; x+2y=32 \Rightarrow 1+2y=32 \Rightarrow y=15.5 \Rightarrow (1,15.5)
\displaystyle y=1,\; x+y=24 \Rightarrow x=23 \Rightarrow (23,1)
\displaystyle \text{Also } (1,1)

\displaystyle \text{The feasible region determined by the constraints is represented graphically.}\displaystyle Z=100x+300y
\displaystyle Z(1,1)=100+300=400
\displaystyle Z(23,1)=2300+300=2600
\displaystyle Z(16,8)=1600+2400=4000
\displaystyle Z(1,15.5)=100+4650=4750
\displaystyle Z_{\max}=4750
\displaystyle \text{Hence the firm should produce }1\text{ necklace and }15.5\text{ bracelets per day for maximum profit.}

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