Question 1.
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:

\displaystyle \begin{array}{|c|c|c|} \hline \textbf{Gadget} & \textbf{Foundry} & \textbf{Machine\!-\!shop} \\ \hline A & 10 & 5 \\ B & 6 & 4 \\ \hline \textbf{Firm's capacity per week} & 1000 & 600 \\ \hline \end{array}

The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ and } y \text{ be the number of gadgets } A \text{ and } B \text{ respectively being} \\ \text{produced in order to maximize the profit.}
\displaystyle \text{Since each unit of gadget } A \text{ takes } 10 \text{ hours to be produced by machine } A \text{ and } 6 \text{ hours} \\ \text{to be produced by machine } B \text{ and each unit of gadget } B \text{ takes } 5 \text{ hours to be} \\ \text{produced by machine } A \text{ and } 4 \text{ hours to be produced by machine } B.
\displaystyle \text{Therefore, the total time taken by the foundry to produce } x \text{ units of gadget } A \text{ and } \\ y \text{ units of gadget } B \text{ is } 10x+6y.
\displaystyle \text{This must be less than or equal to the total hours available.}
\displaystyle 10x+6y\le1000.
\displaystyle \text{This is our first constraint.}
\displaystyle \text{The total time taken by the machine shop to produce } x \text{ units of gadget } A \text{ and } \\ y \text{ units of gadget } B \text{ is } 5x+4y.
\displaystyle \text{This must be less than or equal to the total hours available.}
\displaystyle 5x+4y\le600.
\displaystyle \text{This is our second constraint.}
\displaystyle \text{Since } x \text{ and } y \text{ are non negative integers, therefore}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{It is given that the profit on the sale of } A \text{ is Rs }30 \text{ per unit as compared with Rs }\\ 20 \text{ per unit of } B.
\displaystyle \text{Therefore, profit gained on } x \text{ and } y \text{ number of gadgets } A \text{ and } B \text{ is Rs }30x \text{ and Rs } \\ 20y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=\text{Rs}(30x+20y).
\displaystyle \text{Hence, the above LPP can be stated mathematically as follows:}
\displaystyle \text{Maximize } Z=30x+20y.
\displaystyle \text{Subject to}
\displaystyle 10x+6y\le1000,
\displaystyle 5x+4y\le600,
\displaystyle x\ge0,\;y\ge0.

Question 2.
A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let the company produce } x \text{ units of product } A \text{ and } y \text{ units of product } B.
\displaystyle \text{Since each unit of product } A \text{ costs Rs }60 \text{ and each unit of product } B \text{ costs Rs }80.
\displaystyle \text{Therefore, } x \text{ units of product } A \text{ and } y \text{ units of product } B \text{ will cost Rs }60x \text{ and Rs }80y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote the total cost.}
\displaystyle Z=\text{Rs}(60x+80y).
\displaystyle \text{Also, one unit of product } A \text{ requires one machine hour.}
\displaystyle \text{The total machine hours available with the company for product } A \text{ are } 400 \text{ hours.}
\displaystyle x\le400.
\displaystyle \text{This is our first constraint.}
\displaystyle \text{Also, one unit of product } A \text{ and } B \text{ require } 1 \text{ labour hour each and there are a total of } 500 \\ \text{ labour hours.}
\displaystyle x+y\le500.
\displaystyle \text{This is our second constraint.}
\displaystyle \text{Since } x \text{ and } y \text{ are non-negative integers, therefore}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Also, as per agreement, the company has to supply at least } 200 \text{ units of product } B \text{ to its} \\ \text{regular customers.}
\displaystyle y\ge200.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Minimize } Z=60x+80y.
\displaystyle \text{Subject to } x\le400,
\displaystyle x+y\le500,
\displaystyle y\ge200,
\displaystyle x\ge0,\;y\ge0.

Question 3.
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product.

\displaystyle \begin{array}{|c|c|c|c|} \hline \textbf{Machine} & \textbf{A} & \textbf{B} & \textbf{C} \\ \hline M_{1} & 4 & 3 & 5 \\ M_{2} & 2 & 2 & 4 \\ \hline \end{array}

Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150 A’s. Set up a LPP to maximize the profit.

\displaystyle \text{Answer:}

\displaystyle \text{Let the number of units of products } A,B \text{ and } C \text{ manufactured be } x,y \text{ and } z \text{ respectively.}
\displaystyle \text{Given, machine } M_1 \text{ takes } 4 \text{ minutes to manufacture one unit of product } A,\;3 \text{ minutes to} \\ \text{manufacture one unit of product } B \text{ and } 5 \text{ minutes to manufacture one unit of product } C.
\displaystyle \text{Machine } M_2 \text{ takes } 2 \text{ minutes to manufacture one unit of product } A,\;2 \text{ minutes to} \\ \text{manufacture one unit of product } B \text{ and } 4 \text{ minutes to manufacture one unit of product } C.
\displaystyle \text{The availability is } 2000 \text{ minutes for } M_1 \text{ and } 2500 \text{ minutes for } M_2.
\displaystyle \text{Thus,}
\displaystyle 4x+3y+5z\le2000.
\displaystyle 2x+2y+4z\le2500.
\displaystyle \text{The number of units of products cannot be negative.}
\displaystyle \text{So,}
\displaystyle x\ge0,\;y\ge0,\;z\ge0.
\displaystyle \text{Further, it is given that the firm should manufacture } 100 \text{ units of } A,\;200 \text{ units of } B \\ \text{ and } 50 \text{ units of } C \text{ but not more than } 150 \text{ units of } A.
\displaystyle 100\le x\le150.
\displaystyle y\ge200.
\displaystyle z\ge50.
\displaystyle \text{Let } Z \text{ denote the profit.}
\displaystyle Z=3x+2y+4z.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=3x+2y+4z.
\displaystyle \text{subject to}
\displaystyle 4x+3y+5z\le2000,
\displaystyle 2x+2y+4z\le2500,
\displaystyle 100\le x\le150,
\displaystyle y\ge200,
\displaystyle z\ge50,
\displaystyle x\ge0,\;y\ge0,\;z\ge0.

Question 4.
A firm manufactures two types of products A and B and sells them at a profit of Rs 2 on type A and Rs 3 on type B. Each product is processed on two machines M1 and M2. Type A requires one minute of processing time on M1 and two minutes of M2; type B requires one minute on M1 and one minute on M2. The machine M1 is available for not more than 6 hours 40 minutes while machine M2 is available for 10 hours during any working day. Formulate the problem as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let the firm produce } x \text{ units of product } A \text{ and } y \text{ units of product } B.
\displaystyle \text{Since each unit of product } A \text{ requires one minute on machine } M_1 \text{ and two minutes} \\ \text{on machine } M_2.
\displaystyle \text{Therefore, } x \text{ units of product } A \text{ will require } x \text{ minutes on machine } M_1 \text{ and } 2x \text{ minutes} \\ \text{on machine } M_2.
\displaystyle \text{Also,}
\displaystyle \text{Since each unit of product } B \text{ requires one minute on machine } M_1 \text{ and one minute} \\ \text{on machine } M_2.
\displaystyle \text{Therefore, } y \text{ units of product } B \text{ will require } y \text{ minutes on machine } M_1 \text{ and } y \text{ minutes} \\ \text{on machine } M_2.
\displaystyle \text{It is given that machine } M_1 \text{ is available for } 6 \text{ hours and } 40 \text{ minutes, i.e. } 400 \text{ minutes,} \\ \text{and machine } M_2 \text{ is available for } 10 \text{ hours, i.e. } 600 \text{ minutes.}
\displaystyle \text{Thus,}
\displaystyle x+y\le400.
\displaystyle 2x+y\le600.
\displaystyle \text{Since units of the products cannot be negative, so}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=2x+3y.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=2x+3y.
\displaystyle \text{subject to } x+y\le400,
\displaystyle 2x+y\le600,
\displaystyle x\ge0,\;y\ge0.

Question 5.
A rubber company is engaged in producing three types of tyres A, B and C. Each type requires processing in two plants, Plant I and Plant II. The capacities of the two plants, in number of tyres per day, are as follows:

\displaystyle \begin{array}{|c|c|c|c|} \hline \textbf{Plant} & \textbf{A} & \textbf{B} & \textbf{C} \\ \hline \text{I} & 50 & 100 & 100 \\ \text{II} & 60 & 60 & 200 \\ \hline \end{array}

The monthly demand for tyre A, B and C is 2500, 3000 and 7000 respectively. If Plant I costs Rs 2500 per day and Plant II costs Rs 3500 per day to operate, how many days should each be run per month to minimize cost while meeting the demand? Formulate the problem as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let plant I be run for } x \text{ days and plant II be run for } y \text{ days.}
\displaystyle \text{Then,}
\displaystyle \text{The production data are given in the following table:}
\displaystyle \begin{array}{c|c|c|c} \text{Tyres} & \text{Plant I }(x) & \text{Plant II }(y) & \text{Demand}\\ \hline A & 50 & 60 & 2500\\ B & 100 & 60 & 3000\\ C & 100 & 200 & 7000 \end{array}
\displaystyle \text{Minimum demand for tyres } A,B \text{ and } C \text{ is } 2500,3000 \text{ and } 7000 \text{ respectively.}
\displaystyle \text{The demand can be more than the minimum demand.}
\displaystyle \text{Therefore, the inequalities will be}
\displaystyle 50x+60y\ge2500.
\displaystyle 100x+60y\ge3000.
\displaystyle 100x+200y\ge7000.
\displaystyle \text{Also, the objective function is } Z=2500x+3500y.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Minimize } Z=2500x+3500y.
\displaystyle \text{subject to}
\displaystyle 50x+60y\ge2500,
\displaystyle 100x+60y\ge3000,
\displaystyle 100x+200y\ge7000.

Question 6.
A company sells two different products A and B. The two products are produced in a common production process and are sold in two different markets. The production process has a total capacity of 45000 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 company officials feel that the maximum number of units of A that can be sold is 7000 and that of B is 10000. If the profit is Rs 60 per unit for product A and Rs 40 per unit for product B, how many units of each product should be sold to maximize profit? Formulate the problem as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ units of product } A \text{ and } y \text{ units of product } B \text{ be produced.}
\displaystyle \text{Then,}
\displaystyle \text{Since it takes } 5 \text{ hours to produce one unit of } A \text{ and } 3 \text{ hours to produce one unit of } B.
\displaystyle \text{Therefore, it will take } 5x \text{ hours to produce } x \text{ units of } A \text{ and } 3y \text{ hours to produce } y \text{ units of } B.
\displaystyle \text{As the total capacity is } 45000 \text{ man-hours,}
\displaystyle 5x+3y\le45000.
\displaystyle \text{Also,}
\displaystyle \text{The maximum number of units of } A \text{ that can be sold is } 7000 \text{ and that of } B \text{ is } 10000, \\ \text{ and the number ofunits cannot be negative.}
\displaystyle 0\le x\le7000,\;0\le y\le10000.
\displaystyle \text{Now,}
\displaystyle \text{Total profit }=60x+40y.
\displaystyle \text{Here, we need to maximize profit.}
\displaystyle \text{Thus, the objective function will be}
\displaystyle Z=60x+40y.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=60x+40y.
\displaystyle \text{subject to}
\displaystyle 5x+3y\le45000,
\displaystyle x\le7000,
\displaystyle y\le10000.

Question 7.
To maintain his health a person must fulfill certain minimum daily requirements for several kinds of nutrients. Assuming there are only three kinds of nutrients — calcium, protein and calories — and the person’s diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:

\displaystyle \begin{array}{|c|c|c|c|} \hline & \textbf{Food I (per lb)} & \textbf{Food II (per lb)} & \textbf{Minimum daily requirement} \\ \hline \text{Calcium} & 10 & 5 & 20 \\ \text{Protein} & 5 & 4 & 20 \\ \text{Calories} & 2 & 6 & 13 \\ \hline \textbf{Price (Rs)} & 60 & 100 & \\ \hline \end{array}

What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let the person take } x \text{ lbs and } y \text{ lbs of food I and II respectively in the diet.}
\displaystyle \text{Since per lb of food I costs Rs }60 \text{ and that of food II costs Rs }100.
\displaystyle \text{Therefore, } x \text{ lbs of food I cost Rs }60x \text{ and } y \text{ lbs of food II cost Rs }100y.
\displaystyle \text{Total cost per day }=\text{Rs}(60x+100y).
\displaystyle \text{Let } Z \text{ denote the total cost per day.}
\displaystyle Z=60x+100y.
\displaystyle \text{Total amount of calcium in the diet is } 10x+5y.
\displaystyle \text{Since each lb of food I contains } 10 \text{ units of calcium, therefore } x \text{ lbs of food I contain } 10x \text{ units of calcium.}
\displaystyle \text{Each lb of food II contains } 5 \text{ units of calcium, so } y \text{ lbs of food II contain } 5y \text{ units of calcium.}
\displaystyle \text{Thus, } x \text{ lbs of food I and } y \text{ lbs of food II contain } 10x+5y \text{ units of calcium.}
\displaystyle \text{But the minimum requirement is } 20 \text{ units of calcium.}
\displaystyle 10x+5y\ge20.
\displaystyle \text{Since each lb of food I contains } 5 \text{ units of protein, therefore } x \text{ lbs of food I contain } 5x \text{ units of protein.}
\displaystyle \text{Each lb of food II contains } 4 \text{ units of protein, so } y \text{ lbs of food II contain } 4y \text{ units of protein.}
\displaystyle \text{Thus, } x \text{ lbs of food I and } y \text{ lbs of food II contain } 5x+4y \text{ units of protein.}
\displaystyle \text{But the minimum requirement is } 20 \text{ units of protein.}
\displaystyle 5x+4y\ge20.
\displaystyle \text{Since each lb of food I contains } 2 \text{ units of calories, therefore } x \text{ lbs of food I contain } 2x \text{ units of calories.}
\displaystyle \text{Each lb of food II contains } 6 \text{ units of calories, so } y \text{ lbs of food II contain } 6y \text{ units of calories.}
\displaystyle \text{Thus, } x \text{ lbs of food I and } y \text{ lbs of food II contain } 2x+6y \text{ units of calories.}
\displaystyle \text{But the minimum requirement is } 13 \text{ units of calories.}
\displaystyle 2x+6y\ge13.
\displaystyle \text{Finally, the quantities of food I and food II are non-negative values.}
\displaystyle \text{So,}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Minimize } Z=60x+100y.
\displaystyle \text{subject to}
\displaystyle 10x+5y\ge20,
\displaystyle 5x+4y\ge20,
\displaystyle 2x+6y\ge13,
\displaystyle x\ge0,\;y\ge0.

Question 8.
A manufacturer can produce two products, A and B, during a given time period. Each of these products requires four different manufacturing operations: grinding, turning, assembling and testing. The manufacturing requirements in hours per unit of products A and B are given below:

\displaystyle \begin{array}{|c|c|c|} \hline \textbf{Operation} & \textbf{A} & \textbf{B} \\ \hline \text{Grinding} & 1 & 2 \\ \text{Turning} & 3 & 1 \\ \text{Assembling} & 6 & 3 \\ \text{Testing} & 5 & 4 \\ \hline \end{array}

The available capacities of these operations in hours for the given time period are: grinding 30; turning 60; assembling 200; testing 200. The contribution to profit is Rs 20 for each unit of A and Rs 30 for each unit of B. The firm can sell all that it produces at the prevailing market price. Determine the optimum amount of A and B to produce during the given time period. Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ and } y \text{ units of products } A \text{ and } B \text{ be manufactured respectively.}
\displaystyle \text{The contribution to profit is Rs }2 \text{ for each unit of } A \text{ and Rs }3 \text{ for each unit of } B.
\displaystyle \text{Therefore, for } x \text{ units of } A \text{ and } y \text{ units of } B,\text{ the contribution to profit would be} \\ \text{Rs }2x \text{ and Rs }3y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=\text{Rs}(2x+3y).
\displaystyle \text{Total hours required for grinding, turning, assembling and testing are } x+2y,\;3x+y,\;6x+3y,\;5x+4y \text{ respectively.}
\displaystyle \text{The available capacities of these operations in hours for the given period are grinding } 30, \\ \text{ turning } 60,\text{ assembling } 200 \text{ and testing } 200.
\displaystyle x+2y\le30.
\displaystyle 3x+y\le60.
\displaystyle 6x+3y\le200.
\displaystyle 5x+4y\le200.
\displaystyle \text{Units of products cannot be negative. Therefore,}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=2x+3y.
\displaystyle \text{subject to}
\displaystyle x+2y\le30,
\displaystyle 3x+y\le60,
\displaystyle 6x+3y\le200,
\displaystyle 5x+4y\le200,
\displaystyle x\ge0,\;y\ge0.

Question 9.
Vitamins A and B are found in two different foods F1 and F2. One unit of food F1 contains 2 units of vitamin A and 3 units of vitamin B. One unit of food F2 contains 4 units of vitamin A and 2 units of vitamin B. One unit of food F1 and F2 cost Rs 50 and 25 respectively. The minimum daily requirements for a person of vitamin A and B is 40 and 50 units respectively. Assuming that anything in excess of daily minimum requirement of vitamin A and B is not harmful, find out the optimum mixture of food F1 and F2 at the minimum cost which meets the daily minimum requirement of vitamin A and B. Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ and } y \text{ units of food } F_1 \text{ and } F_2 \text{ be mixed.}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{One unit of food } F_1 \text{ contains } 2 \text{ units of vitamin } A \text{ and one unit of food } F_2 \text{ contains } 4 \text{ units of vitamin } A.
\displaystyle \text{Therefore, } x \text{ and } y \text{ units of food } F_1 \text{ and } F_2 \text{ respectively contain } 2x \text{ and } 4y \text{ units of vitamin } A.
\displaystyle \text{It is given that the minimum daily requirement for a person of vitamin } A \text{ is } 40 \text{ units.}
\displaystyle 2x+4y\ge40.
\displaystyle \text{One unit of food } F_1 \text{ contains } 3 \text{ units of vitamin } B \text{ and one unit of food } F_2 \text{ contains } 2 \text{ units of vitamin } B.
\displaystyle \text{Therefore, } x \text{ and } y \text{ units of } F_1 \text{ and } F_2 \text{ respectively contain } 3x \text{ and } 2y \text{ units of vitamin } B.
\displaystyle \text{It is given that the minimum daily requirement for a person of vitamin } B \text{ is } 50 \text{ units.}
\displaystyle 3x+2y\ge50.
\displaystyle \text{One unit of food } F_1 \text{ and food } F_2 \text{ cost Rs }50 \text{ and } 25 \text{ respectively.}
\displaystyle \text{Therefore, } x \text{ and } y \text{ units of food } F_1 \text{ and food } F_2 \text{ cost Rs }50x \text{ and Rs }25y \text{ respectively.}
\displaystyle \text{Let } Z \text{ denote the total cost.}
\displaystyle Z=\text{Rs}(50x+25y).
\displaystyle \text{Hence, the required LPP is}
\displaystyle \text{Minimize } Z=50x+25y.
\displaystyle \text{subject to}
\displaystyle 2x+4y\ge40,
\displaystyle 3x+2y\ge50,
\displaystyle x\ge0,\;y\ge0.

Question 10.
An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile. Shop B, which performs finishing operations, must work 3 man-days for each automobile or truck it produces. Because of men and machine limitations, shop A has 180 man-days per week available while shop B has 135 man-days per week. If the manufacturer makes a profit of Rs 30000 on each truck and Rs 20000 on each automobile, how many of each should he produce to maximize his profit? Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ number of trucks and } y \text{ number of automobiles be produced to maximize the profit.}
\displaystyle \text{Since the manufacturer makes a profit of Rs }30000 \text{ on each truck and Rs }2000 \text{ on each automobile.}
\displaystyle \text{Therefore, on } x \text{ trucks and } y \text{ automobiles, profit would be Rs }30000x \text{ and Rs }2000y \text{ respectively.}
\displaystyle \text{Total profit }=\text{Rs}(30000x+2000y).
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=30000x+2000y.
\displaystyle \text{Since } 5 \text{ man-days and } 2 \text{ man-days are required to produce each truck and automobile at shop } A.
\displaystyle \text{Therefore, } 5x \text{ man-days and } 2y \text{ man-days are required to produce } x \text{ trucks and } y \text{ automobiles at shop } A.
\displaystyle \text{Also,}
\displaystyle \text{Since } 3 \text{ man-days are required to produce each truck and automobile at shop } B.
\displaystyle \text{Therefore, } 3x \text{ man-days and } 3y \text{ man-days are required to produce } x \text{ trucks and } y \text{ automobiles at shop } B.
\displaystyle \text{As shop } A \text{ has } 180 \text{ man-days per week available while shop } B \text{ has } 135 \text{ man-days per week.}
\displaystyle 5x+2y\le180.
\displaystyle 3x+3y\le135.
\displaystyle \text{Number of trucks and automobiles cannot be negative.}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=30000x+2000y.
\displaystyle \text{subject to}
\displaystyle 5x+2y\le180,
\displaystyle 3x+3y\le135,
\displaystyle x\ge0,\;y\ge0.

Question 11.
Two tailors A and B earn Rs 150 and Rs 200 per day respectively. A can stitch 6 shirts and 4 pants per day while B can stitch 10 shirts and 4 pants per day. For linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 pants. (CBSE 2005)

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ and } y \text{ units of product } A \text{ and } B \text{ be manufactured respectively.}
\displaystyle \text{Labour cost per unit to manufacture product } A \text{ and product } B \text{ is Rs }16 \text{ and} \\ \text{Rs }20 \text{ respectively.}
\displaystyle \text{Therefore, labour cost for } x \text{ and } y \text{ units of product } A \text{ and product } B \text{ is Rs } \\ 16x \text{ and Rs }20y \text{ respectively.}
\displaystyle \text{Total labour cost to manufacture product } A \text{ and product } B \text{ is Rs}(16x+20y).
\displaystyle \text{Raw material cost per unit to manufacture product } A \text{ and product } B \text{ is Rs }4  \\ \text{ and Rs }4 \text{ respectively.}
\displaystyle \text{Therefore, raw material cost for } x \text{ and } y \text{ units of product } A \text{ and product } B \\ \text{ is Rs }4x \text{ and Rs }4y \text{ respectively.}
\displaystyle \text{Total raw material cost to manufacture product } A \text{ and product } B \text{ is Rs}(4x+4y).
\displaystyle \text{Hence, total cost price to manufacture product } A \text{ and product } B=\text{Total labour cost}+\text{Total raw material cost.}
\displaystyle =16x+4x+20y+4y.
\displaystyle =20x+24y.
\displaystyle \text{Selling price per unit for product } A \text{ and product } B \text{ is Rs }25 \text{ and Rs }30 \text{ respectively.}
\displaystyle \text{Therefore, total selling price for product } A \text{ and product } B \text{ is Rs }25x \text{ and Rs }30y \text{ respectively.}
\displaystyle \text{Total selling price}=25x+30y.
\displaystyle \text{Total profit}=\text{Total selling price}-\text{Total cost price}.
\displaystyle =(25x+30y)-(20x+24y).
\displaystyle =5x+6y.
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=5x+6y.
\displaystyle \text{One unit of product } A \text{ and product } B \text{ requires } 3 \text{ hours and } 2 \text{ hours respectively at department } 1.
\displaystyle \text{Therefore, } x \text{ and } y \text{ units of product } A \text{ and product } B \text{ require } 3x \text{ hours and } 2y \text{ hours respectively.}
\displaystyle \text{The weekly capacity of department } 1 \text{ is } 130.
\displaystyle 3x+2y\le130.
\displaystyle \text{One unit of product } A \text{ and product } B \text{ requires } 4 \text{ hours and } 6 \text{ hours respectively at department } 2.
\displaystyle \text{Therefore, } x \text{ and } y \text{ units of product } A \text{ and product } B \text{ require } 4x \text{ hours and } 6y \text{ hours respectively.}
\displaystyle \text{The weekly capacity of department } 2 \text{ is } 260.
\displaystyle 4x+6y\le260.
\displaystyle \text{Units of products cannot be negative. Therefore,}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=5x+6y.
\displaystyle \text{subject to}
\displaystyle 3x+2y\le130,
\displaystyle 4x+6y\le260,
\displaystyle x\ge0,\;y\ge0.

Question 12.
An airline agrees to charter planes for a group. The group needs at least 160 first class seats and at least 300 tourist class seats. The airline must use at least two of its model 314 planes which have 20 first class and 30 tourist class seats. The airline will also use some of its model 535 planes which have 20 first class seats and 60 tourist class seats. Each flight of a model 314 plane costs the company Rs 100000 and each flight of a model 535 plane costs Rs 150000. How many of each type of plane should be used to minimize the flight cost? Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ number of model } 314 \text{ planes and } y \text{ number of model }  \\ 535 \text{ planes be used.}
\displaystyle \text{It is given that the cost of one model } 314 \text{ plane is Rs }100000 \text{ and the} \\ \text{cost of one model } 535 \text{ plane is Rs }150000.
\displaystyle \text{Therefore, the cost of } x \text{ model } 314 \text{ planes is Rs }100000x \text{ and} \\ \text{the cost of } y \text{ model } 535 \text{ planes is Rs }150000y.
\displaystyle \text{Total cost price}=100000x+150000y.
\displaystyle \text{Let } Z \text{ denote the total cost.}
\displaystyle Z=100000x+150000y.
\displaystyle \text{Also,}
\displaystyle \text{Each model } 314 \text{ plane has } 20 \text{ first class and } 30 \text{ tourist class seats,} \\ \text{and each model } 535 \text{ plane has } 20 \text{ first class and } 60 \text{ tourist class seats.}
\displaystyle \text{The group needs } 160 \text{ first class seats and } 300 \text{ tourist class seats.}
\displaystyle 20x+20y\ge160.
\displaystyle 30x+60y\ge300.
\displaystyle \text{Number of planes cannot be negative.}
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Minimize } Z=100000x+150000y.
\displaystyle \text{subject to}
\displaystyle 20x+20y\ge160,
\displaystyle 30x+60y\ge300,
\displaystyle x\ge0,\;y\ge0.

Question 13.
Amit’s mathematics teacher has given him three very long lists of problems with the instruction to submit in time 100 of them correctly solved for credit. The problems in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5-point problem, 2 minutes to solve a 4-point problem, and 4 minutes to solve a 6-point problem. Because he has other subjects to worry about, he cannot afford to devote more than 3½ hours altogether to his mathematics assignment.

Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than 2½ hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{Let Amit correctly solve } x \text{ problems from the first set, } y \text{ problems from the second} \\ \text{set and } z \text{ problems from the third set.}
\displaystyle \text{It is given that Amit cannot submit more than } 100 \text{ correctly solved problems.}
\displaystyle x+y+z\le100.
\displaystyle \text{The problems in the first set are worth } 5 \text{ points each, those in the second set are worth } \\ 4 \text{ points each and those in the third set are worth } 6 \text{ points each.}
\displaystyle \text{Therefore, } x \text{ problems from the first set are worth } 5x \text{ points, } \\ y \text{ problems from the second set are worth } 4y \text{ points and } z \text{ problems from} \\ \text{the third set are worth } 6z \text{ points.}
\displaystyle \text{Thus, total credit points will be } 5x+4y+6z.
\displaystyle \text{Let } Z \text{ denote the total credit of Amit.}
\displaystyle Z=5x+4y+6z.
\displaystyle \text{It requires } 3 \text{ minutes to solve a } 5 \text{-point problem, } 2 \text{ minutes to solve a } \\ 4 \text{-point problem and } 4 \text{ minutes to solve a } 6 \text{-point problem.}
\displaystyle \text{Therefore, } x \text{ problems from the first set require } 3x \text{ minutes, } \\ y \text{ problems from the second set require } 2y \text{ minutes and } z \text{ problems} \\ \text{from the third set require } 4z \text{ minutes.}
\displaystyle \text{Thus, the total time required by Amit will be } (3x+2y+4z) \text{ minutes.}
\displaystyle \text{It is given that the total time that Amit can devote to his mathematics assignment is } \\ 3\frac{1}{2} \text{ hours, i.e. } 210 \text{ minutes.}
\displaystyle 3x+2y+4z\le210.
\displaystyle \text{Further, it is given that the total time that Amit can devote to solving the first two types of} \\ \text{problems cannot be more than } 2\frac{1}{2} \text{ hours, i.e. } 150 \text{ minutes.}
\displaystyle 3x+2y\le150.
\displaystyle \text{Number of problems cannot be negative. Therefore,}
\displaystyle x\ge0,\;y\ge0,\;z\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=5x+4y+6z.
\displaystyle \text{subject to}
\displaystyle x+y+z\le100,
\displaystyle 3x+2y+4z\le210,
\displaystyle 3x+2y\le150,
\displaystyle x\ge0,\;y\ge0,\;z\ge0.

Question 14.
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms for tomatoes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20 per man-day. Formulate this problem as a LPP to maximize the farmer’s total profit.

\displaystyle \text{Answer:}

\displaystyle \text{Let the farmer sow tomatoes in } x \text{ acres, lettuce in } y \text{ acres and radishes in } z \text{ acres} \\ \text{of the farm.}
\displaystyle \text{Average yield per acre is } 2000 \text{ kg for tomatoes, } 3000 \text{ kg for lettuce and } 1000 \text{ kg} \\ \text{for radishes.}
\displaystyle \text{Thus, the farmer raises } 2000x \text{ kg of tomatoes, } 3000y \text{ kg of lettuce and } 1000z \text{ kg of} \\ \text{radishes.}
\displaystyle \text{It is given that the price obtained is Re }1 \text{ per kg for tomatoes, Re }0.75 \text{ per kg for lettuce and} \\ \text{Rs }2 \text{ per kg for radishes.}
\displaystyle \text{Selling price}=\text{Rs}(2000x+2250y+2000z).
\displaystyle \text{Labour required for sowing, cultivating and harvesting per acre is } 5 \text{ man-days for tomatoes} \\ \text{and radishes and } 6 \text{ man-days for lettuce.}
\displaystyle \text{Therefore, labour required is } 5x \text{ for tomatoes, } 6y \text{ for lettuce and } 5z \text{ for radishes.}
\displaystyle \text{Total man-days required}=5x+6y+5z.
\displaystyle \text{Price of one man-day is Rs }20.
\displaystyle \text{Labour cost}=20(5x+6y+5z)=100x+120y+100z.
\displaystyle \text{Also, fertilizer is available at Re }0.50 \text{ per kg and the amount required per acre is } \\ 100 \text{ kg each for tomatoes and lettuce and } 50 \text{ kg for radishes.}
\displaystyle \text{Therefore, fertilizer required is } 100x \text{ kg for tomatoes, } 100y \text{ kg for lettuce and } \\ 50z \text{ kg for radishes.}
\displaystyle \text{Hence, total fertilizer used}=100x+100y+50z.
\displaystyle \text{Fertilizer cost}=\text{Rs }0.5(100x+100y+50z)=\text{Rs}(50x+50y+25z).
\displaystyle \text{So, the total cost to the farmer}=\text{Labour cost}+\text{Fertilizer cost}.
\displaystyle =150x+170y+125z.
\displaystyle \text{Profit made by the farmer}=\text{Selling price}-\text{Cost price}.
\displaystyle =(2000x+2250y+2000z)-(150x+170y+125z).
\displaystyle =1850x+2080y+1875z.
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=1850x+2080y+1875z.
\displaystyle \text{Now, total area of the farm is } 100 \text{ acres.}
\displaystyle x+y+z\le100.
\displaystyle \text{Also, it is given that the total man-days available are } 400.
\displaystyle 5x+6y+5z\le400.
\displaystyle \text{Area of land cannot be negative.}
\displaystyle x\ge0,\;y\ge0,\;z\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=1850x+2080y+1875z.
\displaystyle \text{subject to}
\displaystyle x+y+z\le100,
\displaystyle 5x+6y+5z\le400,
\displaystyle x\ge0,\;y\ge0,\;z\ge0.

Question 15.
A firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost of engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost. (CBSE 2017)

\displaystyle \text{Answer:}

\displaystyle \text{Let the number of large vans and small vans used for transporting the packages be } \\ x \text{ and } y \text{ respectively.}
\displaystyle \text{It is given that the cost of engaging each large van is Rs }400 \text{ and each small van is} \\ \text{Rs }200.
\displaystyle \text{Cost of engaging } x \text{ large vans is Rs }400x.
\displaystyle \text{Cost of engaging } y \text{ small vans is Rs }200y.
\displaystyle \text{Let } Z \text{ be the total cost of engaging } x \text{ large vans and } y \text{ small vans.}
\displaystyle Z=\text{Rs}(400x+200y).
\displaystyle \text{The firm has to transport at least } 1200 \text{ packages daily using large vans which carry } \\ 200 \text{ packages each and small vans which can take } 80 \text{ packages each.}
\displaystyle \text{Number of packages transported by } x \text{ large vehicles plus number of packages} \\ \text{transported by } y \text{ small vehicles is at least } 1200.
\displaystyle 200x+80y\ge1200.
\displaystyle \text{Not more than Rs }3000 \text{ is to be spent daily on transportation.}
\displaystyle 400x+200y\le3000.
\displaystyle \text{Also, the number of large vans cannot exceed the number of small vans.}
\displaystyle x\le y.
\displaystyle \text{Thus, the linear programming problem of the given problem is}
\displaystyle \text{Minimize } Z=400x+200y.
\displaystyle \text{subject to}
\displaystyle 200x+80y\ge1200,
\displaystyle 400x+200y\le3000,
\displaystyle x\le y,
\displaystyle x\ge0,\;y\ge0.

Question 16.
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:

\displaystyle \begin{array}{|c|c|c|c|} \hline & \text{Product A} & \textbf{Product B} & \textbf{Weekly capacity} \\ \hline \text{Department 1} & 3 & 2 & 130 \\ \text{Department 2} & 4 & 6 & 260 \\ \hline \text{Selling price per unit} & 25 & 30 & \\ \text{Labour cost per unit} & 16 & 20 & \\ \text{Raw material cost per unit} & 4 & 4 & \\ \hline \end{array}

The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.

\displaystyle \text{Answer:}

\displaystyle \text{The given data can be arranged in the following tabular form:}
\displaystyle \begin{array}{c|c|c|c} & \text{Product }A & \text{Product }B & \text{Weekly capacity}\\ \hline \text{Department }1 & 3 & 2 & 130\\ \text{Department }2 & 4 & 6 & 260 \end{array}
\displaystyle \begin{array}{c|c|c} & \text{Product }A & \text{Product }B\\ \hline \text{Selling price per unit} & \text{Rs }25 & \text{Rs }30\\ \text{Labour cost per unit} & \text{Rs }16 & \text{Rs }20\\ \text{Raw material cost per unit} & \text{Rs }4 & \text{Rs }4 \end{array}
\displaystyle \text{Let } x \text{ be the number of units of product } A.
\displaystyle \text{Let } y \text{ be the number of units of product } B.
\displaystyle \text{Profit per unit}=\text{Selling price per unit}-\text{Labour cost per unit}-\text{Raw material cost per unit.}
\displaystyle \text{Profit on one unit of product } A=25-16-4=\text{Rs }5.
\displaystyle \text{Profit on one unit of product } B=30-20-4=\text{Rs }6.
\displaystyle \text{Therefore, total profit}=5x+6y.
\displaystyle \text{Let } Z \text{ denote the total profit.}
\displaystyle Z=5x+6y.
\displaystyle \text{According to the question,}
\displaystyle 3x+2y\le130.
\displaystyle 4x+6y\le260.
\displaystyle x\ge0,\;y\ge0.
\displaystyle \text{Hence, the required LPP is as follows:}
\displaystyle \text{Maximize } Z=5x+6y.
\displaystyle \text{subject to}
\displaystyle 3x+2y\le130,
\displaystyle 4x+6y\le260,
\displaystyle x\ge0,\;y\ge0.

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