\displaystyle \textbf{Question 1: } \text{Aman has Rs }1500\text{ to purchase rice and wheat for his grocery shop.}
\displaystyle \text{Each sack of rice and wheat costs Rs }180\text{ and Rs }120\text{ respectively. He can store a}
\displaystyle \text{maximum number of }10\text{ bags in his shop. He will earn a profit of Rs }11\text{ per bag}
\displaystyle \text{of rice and Rs }9\text{ per bag of wheat.}
\displaystyle \text{(i) Formulate a linear programming problem to maximise Aman's profit.}
\displaystyle \text{(ii) Calculate the maximum profit.} \qquad \text{ISC 2024}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the number of wheat and rice bags be }x\text{ and }y.
\displaystyle \text{Since, each sack of rice and wheat cost Rs }180\text{ and Rs }120,\text{ Aman has total Rs }1500.
\displaystyle \therefore 120x+180y\leq1500 \qquad \text{(i)}
\displaystyle \text{Also, he can store maximum number of }10\text{ bags.}
\displaystyle \therefore x+y\leq10 \qquad \text{(ii)}
\displaystyle \text{Thus, the linear programming to maximise Aman's profit is}
\displaystyle \text{Maximise, } Z=9x+11y
\displaystyle \text{Subject to the constraints}
\displaystyle 120x+180y\leq1500
\displaystyle x+y\leq10
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{(ii) To calculate the maximum profit, we draw the feasible region using given constraints.}
\displaystyle \text{Feasible region is }OABCO.
\displaystyle \text{Thus, the corner points of feasible region are}
\displaystyle O(0,0),\ A(10,0),\ B(5,5)\text{ and }C\left(0,\frac{25}{3}\right).
\displaystyle \text{The value of }Z\text{ at corner point is}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner\ point} & Z=9x+11y \\ \hline O(0,0) & 0 \\ \hline A(10,0) & 90 \\ \hline B(5,5) & 100\ (\text{Maximum}) \\ \hline C\left(0,\frac{25}{3}\right) & 91.67 \\ \hline \end{array}
\displaystyle \text{The maximum value of }Z\text{ is }100\text{ at point }(5,5).
\displaystyle \therefore \text{The maximum profit is Rs }100.
\\

\displaystyle \textbf{Question 2: } \text{Solve the following linear programming problem graphically.} \qquad \text{ISC 2023}
\displaystyle \text{Maximise } Z=5x+2y \text{ subject to}
\displaystyle x-2y\leq2,\;3x+2y\leq12,\;-3x+2y\leq3,\;x\geq0,\;y\geq0
\displaystyle \text{Answer:}
\displaystyle \text{We have, Maximize } Z=5x+2y
\displaystyle \text{Subject to constraints,}
\displaystyle x-2y\leq2 \qquad \text{(i)}
\displaystyle 3x+2y\leq12 \qquad \text{(ii)}
\displaystyle -3x+2y\leq3 \qquad \text{(iii)}
\displaystyle x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Given constraints can be written in equation form as}
\displaystyle x-2y=2 \qquad \text{(v)}
\displaystyle 3x+2y=12 \qquad \text{(vi)}
\displaystyle -3x+2y=3 \qquad \text{(vii)}
\displaystyle x=0,\ y=0 \qquad \text{(viii)}
\displaystyle \text{Table for line }x-2y=2
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 2 \\ \hline y & -1 & 0 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x-2y\leq2,\text{ we get}
\displaystyle 0-0\leq2,\text{ which is true.}
\displaystyle \text{So, the half plane is toward the origin.}
\displaystyle \text{Table for line }3x+2y=12
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 4 \\ \hline y & 6 & 0 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+2y\leq12,\text{ we get}
\displaystyle 0+0\leq12,\text{ which is true.}
\displaystyle \text{So, the half plane is toward the origin.}
\displaystyle \text{Table for line }-3x+2y=3
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & -1 \\ \hline y & \frac{3}{2} & 0 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }-3x+2y\leq3,\text{ we get}
\displaystyle 0+0\leq3,\text{ which is true.}
\displaystyle \text{So, the half plane is toward the origin.}
\displaystyle \text{On drawing graph, we have}
\displaystyle \text{The corner points of the feasible region are}
\displaystyle O(0,0),\ A(2,0),\ B\left(\frac{7}{2},\frac{3}{4}\right),\ C\left(\frac{3}{2},\frac{15}{4}\right)\text{ and }D\left(0,\frac{3}{2}\right).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner\ points} & Z=5x+2y \\ \hline O(0,0) & 0 \\ \hline A(2,0) & 10 \\ \hline B\left(\frac{7}{2},\frac{3}{4}\right) & 19\ (\text{Maximum}) \\ \hline C\left(\frac{3}{2},\frac{15}{4}\right) & 15 \\ \hline D\left(0,\frac{3}{2}\right) & 3 \\ \hline \end{array}
\displaystyle \text{Hence, maximum value of }Z\text{ is }19\text{ at }B\left(\frac{7}{2},\frac{3}{4}\right).
\\

\displaystyle \textbf{Question 3: } \text{A company uses three machines to manufacture two types of shirts, half }
\displaystyle \text{sleeves and full sleeves. The number of hours required per week on machine }M_1,\;M_2\text{ and }M_3
\displaystyle \text{for one shirt of each type is given in the following table.}
\displaystyle \begin{array}{|c|c|c|c|} \hline & M_1 & M_2 & M_3 \\ \hline \text{Half sleeves} & 1 & 2 & \frac{8}{5} \\ \hline \text{Full sleeves} & 2 & 1 & \frac{8}{5} \\ \hline \end{array}
\displaystyle \text{None of the machines can be in operations for more than }40\text{ h/week. The profit on each}
\displaystyle \text{half sleeve shirt is Rs }1\text{ and the profit on each full sleeve shirt is Rs }1.50.\text{ How many of }
\displaystyle \text{each type of shirts should be made per week to maximise the company's profit?} \qquad \text{ISC 2020}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the number of half sleeves shirts }=x
\displaystyle \text{and the number of full sleeves shirts }=y
\displaystyle \text{Maximise profit } Z=x+1.5y \qquad \text{(i)}
\displaystyle \text{Subject to the constraints}
\displaystyle x+2y\leq40 \qquad \text{(ii)}
\displaystyle 2x+y\leq40 \qquad \text{(iii)}
\displaystyle \frac{8}{5}x+\frac{8}{5}y\leq40
\displaystyle \Rightarrow x+y\leq25 \qquad \text{(iv)}
\displaystyle x\geq0,\;y\geq0
\displaystyle \text{Now, table for the line }x+2y=40\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 40 \\ \hline y & 20 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,20)\text{ and }(40,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq40,\text{ we get}
\displaystyle 0+0\leq40,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin.}
\displaystyle \text{Table for the line }2x+y=40\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 20 \\ \hline y & 40 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,40)\text{ and }(20,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+y\leq40,\text{ we get}
\displaystyle 0+0\leq40,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin.}
\displaystyle \text{Now, table for the line }x+y=25\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 25 \\ \hline y & 25 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,25)\text{ and }(25,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq25,\text{ we get}
\displaystyle 0+0\leq25,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin. Also, }x\geq0,\;y\geq0
\displaystyle \text{implies that the feasible region lies in the first quadrant.}
\displaystyle \text{Now, draw the lines }x+2y=40,\;2x+y=40\text{ and }x+y=25.
\displaystyle \text{The point of intersection of the lines }x+2y=40\text{ and }x+y=25\text{ is }B(10,15).
\displaystyle \text{The point of intersection of the lines }x+y=25\text{ and }2x+y=40\text{ is }C(15,10).
\displaystyle \text{The shaded region }OABCDO\text{ represents the feasible region and its corner points are}
\displaystyle O(0,0),\;A(0,20),\;B(10,15),\;C(15,10)\text{ and }D(20,0).
\displaystyle \text{The values of }Z\text{ at the corner points are given below:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=x+1.5y \\ \hline O(0,0) & 0+1.5\times0=0 \\ \hline A(0,20) & 0+1.5\times20=30 \\ \hline B(10,15) & 10+1.5\times15=32.5\;(\text{maximum}) \\ \hline C(15,10) & 15+1.5\times10=30 \\ \hline D(20,0) & 20+1.5\times0=20 \\ \hline \end{array}
\displaystyle \text{Here, the maximum value of }Z=32.5\text{ at }x=10,\;y=15.
\displaystyle \therefore \text{Number of half sleeves shirts }=10
\displaystyle \text{and number of full sleeves shirts }=15.
\\

\displaystyle \textbf{Question 4: } \text{A carpenter has }90,\;80\text{ and }50\text{ running feet respectively, of teak wood, plywood}
\displaystyle \text{and rosewood which is used to produce product }A\text{ and product }B.\text{ Each unit of product }A
\displaystyle \text{requires }2,\;1\text{ and }1\text{ running feet and each unit of product }B\text{ requires }1,\;2\text{ and }1
\displaystyle \text{running feet of teak wood, plywood and rosewood, respectively. If product }A\text{ is sold for}
\displaystyle \text{Rs }48\text{ per unit and product }B\text{ is sold for Rs }40\text{ per unit, how many units of product}
\displaystyle A\text{ and product }B\text{ should be produced and sold by the carpenter, in order to obtain the}
\displaystyle \text{maximum gross income?}
\displaystyle \text{Formulate the above as a linear programming problem and solve it, indicating clearly the feasible}
\displaystyle \text{region in the graph.} \qquad \text{ISC 2019}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ units of product }A\text{ and }y\text{ units of product }B\text{ be produced.}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Type} & \text{Number} & \text{Teak wood} & \text{Plywood} & \text{Rosewood} \\ \hline A & x & 2x & x & x \\ \hline B & y & y & 2y & y \\ \hline \end{array}
\displaystyle \therefore \text{The objective function is to maximise}
\displaystyle Z=48x+40y
\displaystyle \text{Subject to the constraints}
\displaystyle 2x+y\leq90,\;x+2y\leq80,\;x+y\leq50,\;x\geq0,\;y\geq0
\displaystyle \text{Consider the constraints as linear equations:}
\displaystyle 2x+y=90 \qquad \text{(i)}
\displaystyle x+2y=80 \qquad \text{(ii)}
\displaystyle x+y=50 \qquad \text{(iii)}
\displaystyle x=0,\;y=0 \qquad \text{(iv)}
\displaystyle \text{Table for the line }2x+y=90\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 45 \\ \hline y & 90 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,90)\text{ and }(45,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+y\leq90,\text{ we get}
\displaystyle 0+0\leq90,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin.}
\displaystyle \text{Table for the line }x+2y=80\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 80 \\ \hline y & 40 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,40)\text{ and }(80,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq80,\text{ we get}
\displaystyle 0+0\leq80,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin.}
\displaystyle \text{Table for the line }x+y=50\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 50 \\ \hline y & 50 & 0 \\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,50)\text{ and }(50,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq50,\text{ we get}
\displaystyle 0+0\leq50,\text{ which is true.}
\displaystyle \text{So, the half-plane is towards the origin.}
\displaystyle \text{Also, }x\geq0,\;y\geq0\text{ implies that the feasible region lies in the first quadrant.}
\displaystyle \text{Now, the points of intersection of lines (i) and (iii), and (ii) and (iii),}
\displaystyle \text{are respectively }C(40,10)\text{ and }B(20,30).
\displaystyle \text{The graphical representation of these lines is given below.}
\displaystyle \text{The shaded region }OECBDO\text{ represents the feasible region and its corner points are}
\displaystyle O(0,0),\;B(20,30),\;C(40,10),\;D(0,40)\text{ and }E(45,0).
\displaystyle \text{The values of }Z\text{ at the corner points are given below:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=48x+40y \\ \hline O(0,0) & Z=48\times0+40\times0=0 \\ \hline B(20,30) & Z=48\times20+40\times30=2160 \\ \hline C(40,10) & Z=48\times40+40\times10=2320\;(\text{maximum}) \\ \hline D(0,40) & Z=48\times0+40\times40=1600 \\ \hline E(45,0) & Z=48\times45+40\times0=2160 \\ \hline \end{array}
\displaystyle \text{Thus, }Z\text{ is maximum at }x=40,\;y=10\text{ and the maximum value of }Z\text{ is }2320.
\displaystyle \text{Hence, }40\text{ units of product }A\text{ and }10\text{ units of product }B\text{ should be produced}
\displaystyle \text{to obtain the maximum gross income of Rs }2320.
\\

 

\displaystyle \textbf{Question 5: } \text{A manufacturing company makes two types of teaching aids }A\text{ and }B\text{ of}
\displaystyle \text{Mathematics for Class }X.\text{ Each type of }A\text{ requires }9\text{ labour hours for fabricating}
\displaystyle \text{and }1\text{ labour hour for finishing. Each type of }B\text{ requires }12\text{ labour hours for}
\displaystyle \text{fabricating and }3\text{ labour hours for finishing.}
\displaystyle \text{For fabricating and finishing, the maximum labour hours available per week are }180\text{ and}
\displaystyle 30,\text{ respectively.}
\displaystyle \text{The company makes a profit of Rs }80\text{ on each piece of type }A\text{ and Rs }120\text{ on each}
\displaystyle \text{piece of type }B.\text{ How many pieces of type }A\text{ and type }B\text{ should be manufactured}
\displaystyle \text{per week to get a maximum profit? Formulate this as linear programming problem and solve}
\displaystyle \text{it. Identify the feasible region from the rough sketch.} \qquad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the number of pieces of two types of teaching aids}
\displaystyle A\text{ and }B\text{ be }x\text{ and }y,\text{ respectively.}
\displaystyle \text{We construct the following table.}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Items} & \text{Number of pieces} & \text{Time on fabricating (in h)} & \text{Time on finishing (in h)} & \text{Profit in (Rs)} \\ \hline A & x & 9x & x & 80x \\ \hline B & y & 12y & 3y & 120y \\ \hline \text{Total} & x+y & 9x+12y & x+3y & 80x+120y \\ \hline \text{Availability} & & 180 & 30 & \\ \hline \end{array}
\displaystyle \text{The profit on type }A\text{ is Rs }80\text{ and on type }B\text{ is Rs }120.
\displaystyle \text{Our objective is to maximise,}
\displaystyle Z=80x+120y \qquad \text{(i)}
\displaystyle \text{Subject to constraints,}
\displaystyle 9x+12y\leq180\ \text{or}\ 3x+4y\leq60 \qquad \text{(ii)}
\displaystyle x+3y\leq30 \qquad \text{(iii)}
\displaystyle \text{and }x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for line }9x+12y=180\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 20 \\ \hline y & 15 & 0 \\ \hline \end{array}
\displaystyle \text{So, line passes through the points }(0,15)\text{ and }(20,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }9x+12y\leq180,
\displaystyle \text{we get}
\displaystyle 9(0)+12(0)\leq180
\displaystyle \Rightarrow 0\leq180,\text{ which is true.}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }x+3y=30\text{ is}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 30 \\ \hline y & 10 & 0 \\ \hline \end{array}
\displaystyle \text{So, line passes through the points }(0,10)\text{ and }(30,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+3y\leq30,\text{ we get}
\displaystyle \Rightarrow 0\leq30,\text{ which is true.}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ so the region lies in the}
\displaystyle \text{Ist quadrant.}
\displaystyle \text{On solving Eqs. (ii) and (iii), we get the point of}
\displaystyle \text{intersection is }B(12,6).
\displaystyle \text{The graphical representation of the lines is given}
\displaystyle \text{below.}
\displaystyle \text{Feasible region is }OABCO.
\displaystyle \text{The corner points of the feasible region are }O(0,0),
\displaystyle A(20,0),\ B(12,6)\text{ and }C(0,10).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & Z=80x+120y \\ \hline O(0,0) & Z=80(0)+120(0)=0 \\ \hline A(20,0) & Z=80(20)+120(0)=1600 \\ \hline B(12,6) & Z=80(12)+120(6)=1680 \\ \hline C(0,10) & Z=80(0)+120(10)=1200 \\ \hline \end{array}
\displaystyle \text{Hence, the maximum value of }Z\text{ is Rs }1680\text{ at }B(12,6).
\displaystyle \text{Thus, the number of pieces of types }A=12\text{ and}
\displaystyle \text{types }B=6\text{ should be manufactured per week to get a}
\displaystyle \text{maximum profit.}
\\

\displaystyle \textbf{Question 6: } \text{A farmer has a supply of chemical fertilizer of type }A\text{ which contains}
\displaystyle 10\%\text{ nitrogen and }6\%\text{ phosphoric acid and of type }B\text{ which contains }5\%
\displaystyle \text{nitrogen and }10\%\text{ phosphoric acid. After soil test, it is found that at least }7\text{ kg}
\displaystyle \text{of nitrogen and same quantity of phosphoric acid is required for a good crop. The}
\displaystyle \text{fertilizer of type }A\text{ costs Rs }5.00\text{ per kg and the type }B\text{ costs Rs }8.00\text{ per kg.}
\displaystyle \text{Using linear programming, find how many kilograms of each type of the fertilizer should}
\displaystyle \text{be bought to meet the requirement and for the cost to be minimum. Find the feasible}
\displaystyle \text{region in the graph.} \qquad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the farmer uses }x\text{ kg of type }A\text{ fertilizer and }y\text{ kg}
\displaystyle \text{of type }B\text{ fertilizer. Then, we have the following table}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Type} & \text{Quantity (in kg)} & \text{Nitrogen} & \text{Phosphoric acid} & \text{Cost in (Rs)} \\ \hline A & x & \frac{10}{100}x=\frac{1}{10}x & \frac{6}{100}x & 5x \\ \hline B & y & \frac{5}{100}y=\frac{1}{20}y & \frac{10}{100}y & 8y \\ \hline \text{Total} & x+y & \frac{x}{10}+\frac{y}{20} & \frac{6x}{100}+\frac{10y}{100} & 5x+8y \\ \hline \text{Requirement (in kg)} & & 7 & 7 & \\ \hline \end{array}
\displaystyle \text{Total cost of fertilizers }=5x+8y
\displaystyle \text{Since, we have to minimise the cost}
\displaystyle \text{i.e.} \qquad Z=5x+8y \qquad \text{(i)}
\displaystyle \text{Subject to constraints}
\displaystyle \frac{x}{10}+\frac{y}{20}\geq7 \Rightarrow 2x+y\geq140 \qquad \text{(ii)}
\displaystyle \frac{6x}{100}+\frac{10y}{100}\geq7 \Rightarrow 3x+5y\geq350 \qquad \text{(iii)}
\displaystyle \text{and} \qquad x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Firstly, draw the graph of the line}
\displaystyle 2x+y=140
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 70 \\ \hline y & 140 & 0 \\ \hline \end{array}
\displaystyle \text{Putting }(0,0)\text{ in the inequality }2x+y\geq140,\text{ we get}
\displaystyle 2\times0+0\geq140 \Rightarrow 0\geq140,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Since, }x,\ y\geq0
\displaystyle \text{So, the feasible region lies in the first quadrant.}
\displaystyle \text{Secondly, draw the graph of the line}
\displaystyle 3x+5y=350
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & \frac{350}{3} \\ \hline y & 70 & 0 \\ \hline \end{array}
\displaystyle \text{Putting }(0,0)\text{ in the inequality }3x+5y\geq350,\text{ we get}
\displaystyle 3\times0+5\times0\geq350 \Rightarrow 0\geq350,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{On solving the equations }2x+y=140\text{ and}
\displaystyle 3x+5y=350,\text{ we get }B(50,40).
\displaystyle \text{It can be seen that the feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are}
\displaystyle A\left(\frac{350}{3},0\right),\ B(50,40)\text{ and }C(0,140).
\displaystyle \text{The values of }Z\text{ at these points are as follows}
\displaystyle \text{The values of }Z\text{ at the corner points are given}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & Z=5x+8y \\ \hline A\left(\frac{350}{3},0\right) & Z=5\left(\frac{350}{3}\right)+8(0)=\frac{1750}{3}=583 \\ \hline B(50,40) & Z=5(50)+8(40)=570\ (\text{Minimum}) \\ \hline C(0,140) & Z=5(0)+8(140)=1120 \\ \hline \end{array}
\displaystyle \text{As the feasible region is unbounded therefore, }570\text{ may}
\displaystyle \text{or may not be the minimum value of }Z.\text{ For this, we}
\displaystyle \text{draw a graph of the inequality, }5x+8y<570\text{ and check,}
\displaystyle \text{whether the resulting half plane has points in common}
\displaystyle \text{with the feasible region or not.}
\displaystyle \text{It can be seen that the feasible region has no common}
\displaystyle \text{point with }5x+8y<570.
\displaystyle \text{Therefore, }50\text{ kg of fertilizer }A\text{ and }40\text{ kg of fertilizer}
\displaystyle B\text{ should be used to minimize the cost. The minimum}
\displaystyle \text{cost is Rs }570.
\\

\displaystyle \textbf{Question 7: } \text{A company manufactures two types of products }A\text{ and }B.\text{ Each unit of }A
\displaystyle \text{requires }3\text{ g of nickel and }1\text{ g of chromium, while each unit of }B\text{ requires }1\text{ g}
\displaystyle \text{of nickel and }2\text{ g of chromium. The firm can produce }9\text{ g of nickel and }8\text{ g of}
\displaystyle \text{chromium. The profit is Rs }40\text{ on each unit of product of type }A\text{ and Rs }50\text{ on each}
\displaystyle \text{unit of type }B.
\displaystyle \text{How many units of each type should the company manufacture so as to earn maximum}
\displaystyle \text{profit? Use linear programming to find the solution.} \qquad \text{ISC 2016}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the company manufactures }x\text{ units of product }A
\displaystyle \text{and }y\text{ units of product }B.
\displaystyle \text{We construct the following table:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Type} & \text{Unit} & \text{Nickel (in g)} & \text{Chromium (in g)} & \text{Profit} \\ \hline A & x & 3x & x & 40x \\ \hline B & y & y & 2y & 50y \\ \hline \text{Total} & x+y & 3x+y & x+2y & 40x+50y \\ \hline \text{Availability} & & 9 & 8 & \\ \hline \end{array}
\displaystyle \text{Since, the profit is to be maximised}
\displaystyle \text{i.e.} \qquad Z=40x+50y \qquad \text{(i)}
\displaystyle \text{Subject to the constraints are}
\displaystyle 3x+y\leq9 \qquad \text{(ii)}
\displaystyle x+2y\leq8 \qquad \text{(iii)}
\displaystyle \text{and }x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for line }3x+y=9\text{ or }y=9-3x
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 3 \\ \hline y & 9 & 0 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+y\leq9,\text{ we get}
\displaystyle 0+0\leq9 \Rightarrow 0\leq9,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }x+2y=8\text{ or }y=\frac{8-x}{2}
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 8 \\ \hline y & 4 & 0 \\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq8,\text{ we get}
\displaystyle 0+0\leq8 \Rightarrow 0\leq8,\text{ which is true.}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Since, }x,\ y\geq0
\displaystyle \text{So, the feasible region lies in the first quadrant.}
\displaystyle \text{On solving equations }3x+y=9\text{ and }x+2y=8,\text{ we get}
\displaystyle B(2,3).
\displaystyle \text{So, feasible region is }OABCO.
\displaystyle \text{The corner points of the feasible region are }O(0,0),
\displaystyle A(3,0),\ B(2,3)\text{ and }C(0,4).
\displaystyle \text{Now, the values of }Z\text{ at these points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=40x+50y \\ \hline O(0,0) & Z=40(0)+50(0)=0 \\ \hline A(3,0) & Z=40(3)+50(0)=120 \\ \hline B(2,3) & Z=40(2)+50(3)=230\ (\text{Maximum}) \\ \hline C(0,4) & Z=40(0)+50(4)=200 \\ \hline \end{array}
\displaystyle \text{Hence, the maximum value of }Z\text{ is Rs }230\text{ at }B(2,3).
\displaystyle \text{Thus, }2\text{ units of type }A\text{ and }3\text{ units of type }B\text{ should be}
\displaystyle \text{produced, to get the maximum profit of Rs }230.
\\

\displaystyle \textbf{Question 8: } \text{A dietician wishes to mix two kinds of food }X\text{ and }Y\text{ in such a way that}
\displaystyle \text{the mixture contains at least }10\text{ units of vitamin }A,\;12\text{ units of vitamin }B\text{ and}
\displaystyle 8\text{ units of vitamin }C.\text{ The vitamin contents of one kg food is given below:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Food} & \text{Vitamin }A & \text{Vitamin }B & \text{Vitamin }C \\ \hline X & 1\text{ unit} & 2\text{ units} & 3\text{ units} \\ \hline Y & 2\text{ units} & 2\text{ units} & 1\text{ unit} \\ \hline \end{array}
\displaystyle \text{One kg food of }X\text{ costs Rs }24\text{ and one kg food of }Y\text{ costs Rs }36.
\displaystyle \text{Using linear programming, find the least cost of the total mixture which will contain the}
\displaystyle \text{required vitamins.} \qquad \text{ISC 2015}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ kg of food }F_1\text{ and }y\text{ kg of food }F_2\text{ produce the}
\displaystyle \text{required diet. Then, the problem can be formulated as}
\displaystyle \text{an LPP as follows:}
\displaystyle \text{Minimum, } Z=24x+36y \qquad \text{(i)}
\displaystyle \text{Subject to the constraints are}
\displaystyle x+2y\geq10 \qquad \text{(ii)}
\displaystyle 2x+2y\geq12\text{ or }x+y\geq6 \qquad \text{(iii)}
\displaystyle 3x+y\geq8 \qquad \text{(iv)}
\displaystyle \text{and }x\geq0,\ y\geq0 \qquad \text{(v)}
\displaystyle \text{Table for graph of the line }x+2y=10
\displaystyle \begin{array}{|c|c|c|} \hline x & 10 & 0 \\ \hline y & 0 & 5 \\ \hline \end{array}
\displaystyle \text{So, line passes through }(10,0)\text{ and }(0,5).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\geq10,\text{ we get}
\displaystyle 0+0\geq10,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for line }x+y=6.
\displaystyle \begin{array}{|c|c|c|} \hline x & 6 & 0 \\ \hline y & 0 & 6 \\ \hline \end{array}
\displaystyle \text{So, line passes through }(6,0)\text{ and }(0,6).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\geq6,\text{ we get}
\displaystyle 0+0\geq6,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for graph of the line }3x+y=8
\displaystyle \begin{array}{|c|c|c|} \hline x & \frac{8}{3} & 0 \\ \hline y & 0 & 8 \\ \hline \end{array}
\displaystyle \text{So, line passes through }\left(\frac{8}{3},0\right)\text{ and }(0,8).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+y\geq8,\text{ we get}
\displaystyle 0+0\geq8,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Also, }x\geq0,\ y\geq0,\text{ so shaded region lies in Ist quadrant.}
\displaystyle \text{The intersection points of lines (ii), (iii) and (iv), we get}
\displaystyle B(1,5)\text{ and }C(2,4).
\displaystyle \text{The graph is shown below}
\displaystyle \text{The feasible region is }ABCD.
\displaystyle \text{The corner points of the feasible region are}
\displaystyle A(0,8),\ B(1,5),\ C(2,4)\text{ and }D(10,0).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & Z=24x+36y \\ \hline A(0,8) & Z=0+36\times8=288 \\ \hline B(1,5) & Z=24\times1+36\times5 \\ \hline & =24+180=204 \\ \hline C(2,4) & Z=24\times2+36\times4 \\ \hline & =48+144=192\ (\text{least}) \\ \hline D(10,0) & Z=24\times10+36\times0=240 \\ \hline \end{array}
\displaystyle \text{The least value of }Z\text{ is }192\text{ at }C(2,4).\text{ As, the feasible}
\displaystyle \text{region is unbounded, so we draw the graph of the half}
\displaystyle \text{plane }24x+36y<192\text{ and note that there is no point}
\displaystyle \text{common with the feasible region. Therefore, }Z\text{ has}
\displaystyle \text{minimum value and the minimum value is Rs }192.
\\

\displaystyle \textbf{Question 9: } \text{A company manufactures two types of toys }A\text{ and }B.
\displaystyle \text{A toy of type }A\text{ requires }5\text{ min for cutting and }10\text{ min for assembling. A toy of}
\displaystyle \text{type }B\text{ requires }8\text{ min for cutting and }8\text{ min for assembling. There are }3\text{ h}
\displaystyle \text{available for cutting and }4\text{ h available for assembling the toys in a day.}
\displaystyle \text{The profit is Rs }50\text{ each on a toy of type }A\text{ and Rs }60\text{ each on a toy of type }B.
\displaystyle \text{How many toys of each type should company manufacture in a day to maximise the profit?}
\displaystyle \text{Use linear programming to find the solution.} \qquad \text{ISC 2014}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the total number of toys manufactured by the}
\displaystyle \text{company, }A\text{ type to be }x\text{ and }B\text{ type to be }y.
\displaystyle \text{Since, the profit is to be maximised.}
\displaystyle \therefore \text{Maximise profit }(P)=50x+60y \qquad \text{(i)}
\displaystyle \text{So, we can arrange the given data in the following table}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Types of toys} & \text{Total number} & \text{Required time for cutting} & \text{Required time for assembling} & \text{Profit} \\ \hline A & x & 5x\ \text{min} & 10x\ \text{min} & \text{Rs }50x \\ \hline B & y & 8y\ \text{min} & 8y\ \text{min} & \text{Rs }60y \\ \hline \text{Total} & & (5x+8y)\ \text{min} & (10x+8y)\ \text{min} & \text{Rs }(50x+60y) \\ \hline \end{array}
\displaystyle \text{Given, total time for cutting }\leq180\text{ min}
\displaystyle \text{and total time for assembling }\leq240\text{ min}
\displaystyle \therefore 5x+8y\leq180 \qquad \text{(ii)}
\displaystyle \text{and} \qquad 10x+8y\leq240
\displaystyle \text{or} \qquad 5x+4y\leq120 \qquad \text{(iii)}
\displaystyle \text{and} \qquad x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for graph of the line }5x+8y=180.
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 36 \\ \hline y & \frac{45}{2} & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }5x+8y\leq180,\text{ we get}
\displaystyle 0+0\leq180,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for graph of the line }5x+4y=120.
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 24 \\ \hline y & 30 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }5x+4y\leq120,\text{ we get}
\displaystyle 0+0\leq120,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0,\ y\geq0,\text{ so the feasible region lies in the first}
\displaystyle \text{quadrant.}
\displaystyle \text{The point of intersection of Eqs. (ii) and (iii) is }E(12,15).
\displaystyle \text{The shaded region represents the feasible region and}
\displaystyle \text{its corner points are}
\displaystyle O(0,0),\ A\left(0,\frac{45}{2}\right),\ D(24,0)\text{ and }E(12,15).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & P=50x+60y \\ \hline O(0,0) & P=50\times0+60\times0=0 \\ \hline A\left(0,\frac{45}{2}\right) & P=50\times0+60\times\frac{45}{2} \\ \hline & =30\times45=1350 \\ \hline D(24,0) & P=50\times24+60\times0 \\ \hline & =50\times24=1200 \\ \hline E(12,15) & P=50\times12+60\times15 \\ \hline & =600+900=1500 \\ \hline \end{array}
\displaystyle \text{Hence, the profit of the company is maximum of Rs }1500
\displaystyle \text{for manufacturing }12\text{ toys of type }A\text{ and }15\text{ toys of}
\displaystyle \text{type }B.
\\

\displaystyle \textbf{Question 10: } \text{A mill owner buys two types of machines }A\text{ and }B\text{ for his mill. Machine }A
\displaystyle \text{occupies }1000\text{ sq m of area and requires }12\text{ men to operate it, while machine }B
\displaystyle \text{occupies }1200\text{ sq m of area and requires }8\text{ men to operate it. The owner has}
\displaystyle 7600\text{ sq m of area available and }72\text{ men to operate the machines.}
\displaystyle \text{If machine }A\text{ produces }50\text{ units and machine }B\text{ produces }40\text{ units daily, how many}
\displaystyle \text{machines of each type should he buy to maximize the daily output? Use linear programming}
\displaystyle \text{to find the solution.} \qquad \text{ISC 2013}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the number of type }A\text{ machines be }x\text{ and the total}
\displaystyle \text{number of type }B\text{ machines be }y.
\displaystyle \text{Since, the profit is to be maximised.}
\displaystyle \therefore \text{Maximise, } Z=50x+40y \qquad \text{(i)}
\displaystyle \text{Subject to the constraints are}
\displaystyle 1000x+1200y\leq7600\text{ or }5x+6y\leq38 \qquad \text{(ii)}
\displaystyle 12x+8y\leq72\text{ or }3x+2y\leq18 \qquad \text{(iii)}
\displaystyle x\geq0\text{ and }y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for graph of line }5x+6y=38.
\displaystyle \begin{array}{|c|c|c|} \hline x & \frac{38}{5} & 0 \\ \hline y & 0 & \frac{19}{3} \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }5x+6y\leq38,\text{ we get}
\displaystyle 0+0\leq38
\displaystyle \Rightarrow 0\leq38,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for graph of line }3x+2y=18
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 6 \\ \hline y & 9 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }3x+2y\leq18,\text{ we get}
\displaystyle 0+0\leq18
\displaystyle \Rightarrow 0\leq18,\text{ which is true}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0,\ y\geq0,\text{ so the feasible region lies in the first}
\displaystyle \text{quadrant.}
\displaystyle \text{The point of intersection of Eqs. (ii) and (iii) is }B(4,3).
\displaystyle \text{The shaded region represents a bounded feasible}
\displaystyle \text{region and its corner point are }O(0,0),
\displaystyle A(6,0),\ B(4,3),\ C\left(0,\frac{19}{3}\right).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & Z=50x+40y \\ \hline O(0,0) & Z=50\times0+40\times0=0 \\ \hline A(6,0) & Z=50\times6+40\times0=300 \\ \hline B(4,3) & Z=50\times4+40\times3 \\ \hline & =200+120=320 \\ \hline C\left(0,\frac{19}{3}\right) & Z=50\times0+40\times6.3=252 \\ \hline \end{array}
\displaystyle \text{Hence, maximum daily output is }320,\text{ which will be}
\displaystyle \text{achieved when }4\text{ machines of type }A\text{ and }3\text{ machines}
\displaystyle \text{of type }B\text{ are bought.}
\\

\displaystyle \textbf{Question 11: } \text{Two tailors }P\text{ and }Q\text{ earn Rs }150\text{ and Rs }200\text{ per day, respectively.}
\displaystyle P\text{ can stitch }6\text{ shirts and }4\text{ trousers a day, while }Q\text{ can stitch }10\text{ shirts and}
\displaystyle 4\text{ trousers per day. How many days should each work to produce at least }60\text{ shirts}
\displaystyle \text{and }32\text{ trousers at minimum labour cost?} \qquad \text{ISC 2012}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the tailor }P\text{ works for }x\text{ days, while tailor }Q\text{ works}
\displaystyle \text{for }y\text{ days.}
\displaystyle \text{Since, the cost is to be minimised.}
\displaystyle \therefore \text{Minimise }(Z)=150x+200y \qquad \text{(i)}
\displaystyle \text{Subject to the constraints are}
\displaystyle 6x+10y\geq60
\displaystyle \text{or} \qquad 3x+5y\geq30 \qquad \text{(ii)}
\displaystyle 4x+4y\geq32\text{ or }x+y\geq8 \qquad \text{(iii)}
\displaystyle x\geq0,\ y\geq0 \qquad \text{(iv)}
\displaystyle \text{Table for graph of line }3x+5y=30.
\displaystyle \begin{array}{|c|c|c|} \hline x & 10 & 0 \\ \hline y & 6 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }3x+5y\geq30,\text{ we get}
\displaystyle 0+0\geq30 \Rightarrow 0\geq30,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for graph of the line }x+y=8.
\displaystyle \begin{array}{|c|c|c|} \hline x & 0 & 8 \\ \hline y & 8 & 0 \\ \hline \end{array}
\displaystyle \text{Put }(0,0)\text{ in the inequality }x+y\geq8,\text{ we get}
\displaystyle 0+0\geq8 \Rightarrow 0\geq8,\text{ which is false}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Also, }x\geq0,\ y\geq0,\text{ so the feasible region lies in the first}
\displaystyle \text{quadrant. The intersection point of lines (i) and (ii) is}
\displaystyle B(5,3).
\displaystyle \text{The shaded region represents an unbounded feasible}
\displaystyle \text{region and its corner points are }A(10,0),\ B(5,3)\text{ and}
\displaystyle C(0,8).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & Z=150x+200y \\ \hline A(10,0) & Z=150\times10+200\times0=1500 \\ \hline B(5,3) & Z=150\times5+200\times3 \\ \hline & =750+600=1350 \\ \hline C(0,8) & Z=150\times0+200\times8=1600 \\ \hline \end{array}
\displaystyle \text{The cost is minimum at }B(5,3).
\displaystyle \text{But the feasible region is unbounded, therefore }1350
\displaystyle \text{may or may not be the minimum value of }Z.
\displaystyle \text{Now, we draw the graph of the inequality}
\displaystyle 150x+200y<1350,\ \text{i.e.}\ 3x+4y<27\text{ and check}
\displaystyle \text{whether the open half plane has points in common}
\displaystyle \text{with feasible region or not. From the graph, we see}
\displaystyle \text{that the open half plane }3x+4y<27\text{ has no points in}
\displaystyle \text{common with the feasible region.}
\displaystyle \text{So, the minimum value of }Z\text{ is }1350\text{ at }B(5,3).
\displaystyle \text{Hence, tailor }P\text{ should works for }5\text{ days and tailor }Q
\displaystyle \text{should work for }3\text{ days to minimise the labour cost.}
\\

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