\displaystyle \textbf{Question 1:}\\ \text{A dietician wishes to mix two types of food in such a way that the vitamin contents}
\displaystyle \text{of the mixture contain at least 8 units of Vitamin A and 10 units of vitamin C. Food 'I'}
\displaystyle \text{contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C while food 'II'}
\displaystyle \text{contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs Rs.50.00 per}
\displaystyle \text{kg to purchase food 'I' and Rs.70.00 per kg to produce food 'II'. Formulate the above linear}
\displaystyle \text{programming problem to minimize the cost of such a mixture. [CBSE 2011]}

\displaystyle \text{Answer:}
\displaystyle  \text{The given data may be put in the following tabular form:}
\displaystyle \begin{array}{c|cc|c} \text{Resources} & \multicolumn{2}{c|}{\text{Food}} & \text{Requirements}\\ & \text{I} & \text{II} & \\ \hline \text{Vitamin A} & 2 & 1 & 8\\ \text{Vitamin C} & 1 & 2 & 10\\ \text{Cost (in Rs)} & 50 & 70 & \end{array}
\displaystyle \text{Let the dietician mix }x\text{ kg of food 'I' and }y\text{ kg of food 'II'. Clearly, }x \ge 0,\ y \ge 0.
\displaystyle \text{Since one kg of food 'I' costs Rs.50 and one kg of food 'II' costs Rs.70. Therefore,}
\displaystyle \text{total cost of }x\text{ kg of food 'I' and }y\text{ kg of food 'II' is Rs.(50x+70y).}
\displaystyle \text{Let }Z\text{ denote the total cost. Then,}
\displaystyle Z=5x+7y
\displaystyle \text{Since one kg of food 'I' contains 2 units of vitamin A. Therefore, }x\text{ kg of food 'I'}
\displaystyle \text{contain }2x\text{ units of vitamin A. One kg of food 'II' contains one unit of vitamin A.}
\displaystyle \text{So, }y\text{ kg of food 'II' contains }y\text{ units of vitamin A. Thus, }x\text{ kg of food 'I'}
\displaystyle \text{and }y\text{ kg of food 'II' contain }2x+y\text{ units of vitamin A. But, the minimum}
\displaystyle \text{requirement of vitamin A is 8 units.}
\displaystyle 2x+y \ge 8
\displaystyle \text{Similarly, total amount of vitamin C supplied by }x\text{ units of food 'I' and }y\text{ units of}
\displaystyle \text{food 'II' is }(x+2y)\text{ units and the minimum requirement of vitamin C is 10 units.}
\displaystyle x+2y \ge 10
\displaystyle \text{Hence, the mathematical model of the LPP is as follows:}
\displaystyle \text{Minimize }Z=5x+7y
\displaystyle \text{Subject to}
\displaystyle 2x+y \ge 8
\displaystyle x+2y \ge 10
\displaystyle \text{and, }x,\ y \ge 0.

\displaystyle \textbf{Question 2:} \\ \text{A house wife wishes to mix together two kinds of food, X and Y, in such a way}
\displaystyle \text{that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B}
\displaystyle \text{and 8 units of vitamin C. The vitamin contents of one kg of food are given below:}
\displaystyle \begin{array}{c|ccc} & \text{Vitamin A} & \text{Vitamin B} & \text{Vitamin C}\\ \hline \text{Food X} & 1 & 2 & 3\\ \text{Food Y} & 2 & 2 & 1 \end{array}
\displaystyle \text{One kg of food X costs Rs.6 and one kg of food Y costs Rs.10. Find the least cost of}
\displaystyle \text{the mixture which will produce the diet. [CBSE 2003]}

\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ kg of food X and }y\text{ kg of food Y are mixed together to make the mixture.}
\displaystyle \text{Since one kg of food X contains one unit of vitamin A and one kg of food Y contains 2 units}
\displaystyle \text{of vitamin A. Therefore, }x\text{ kg of food X and }y\text{ kg of food Y will contain }x+2y\text{ units of vitamin A.}
\displaystyle \text{But the mixture should contain at least 10 units of vitamin A. Therefore,}
\displaystyle x+2y \ge 10
\displaystyle \text{Similarly, }x\text{ kg of food X and }y\text{ kg of food Y will produce }2x+2y\text{ units of vitamin B}
\displaystyle \text{and }3x+y\text{ units of vitamin C. But the minimum requirements of vitamins B and C are}
\displaystyle \text{respectively of 12 and 8 units.}
\displaystyle 2x+2y \ge 12 \qquad \text{and} \qquad 3x+y \ge 8
\displaystyle \text{Since the quantity of food X and food Y cannot be negative.}
\displaystyle x \ge 0,\ y \ge 0
\displaystyle \text{It is given that one kg of food X costs Rs.6 and one kg of food Y costs Rs.10. So, }x\text{ kg of}
\displaystyle \text{food X and }y\text{ kg of food Y will cost Rs.(6x+10y).}
\displaystyle \text{Thus, the given linear programming problem is}
\displaystyle \text{Minimize }Z=6x+10y
\displaystyle \text{Subject to}
\displaystyle x+2y \ge 10
\displaystyle 2x+2y \ge 12
\displaystyle 3x+y \ge 8
\displaystyle \text{and, }x \ge 0,\ y \ge 0
\displaystyle \text{To solve this LPP, we draw the lines }x+2y=10,\ 2x+2y=12\text{ and }3x+y=8.
\displaystyle \text{The feasible region of the LPP is shaded in Figure below:}\displaystyle \text{The coordinates of the vertices (Corner-points) of shaded feasible region }\text{ are}
\displaystyle (10,0),\ (2,4),\ (1,5)\text{ and }(0,8). \text{These points have been obtained by}
\displaystyle \text{solving the equations of the corresponding intersecting lines, simultaneously.}
\displaystyle \text{The values of the objective function at these points are given in the following table:}
\displaystyle \begin{array}{c|c} text{Point }(x,y) & \text{Value of the objective function }Z=6x+10y\\ \hline (10,0) & Z=6\times10+10\times0=60\\ (2,4) & Z=6\times2+10\times4=52\\ (1,5) & Z=6\times1+10\times5=56\\ (0,8) & Z=6\times0+10\times8=80 \end{array}
\displaystyle \text{Clearly, }Z\text{ is minimum at }x=2\text{ and }y=4.\ \text{The minimum value of }Z\text{ is }52.
\displaystyle \text{We observe that the open half-plane represented by }6x+10y<52\text{ does not have points in common}
\displaystyle \text{with the feasible region. So, }Z\text{ has minimum value equal to }52.
\displaystyle \text{Hence, the least cost of the mixture is Rs.52.}

\displaystyle \textbf{Question 3:} \\  \text{A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture}
\displaystyle \text{contain at least 8 units of vitamin A and 10 units of vitamin C. Food 'I' contains 2 units}
\displaystyle \text{kg of vitamin A and 1 unit kg of vitamin C while food 'II' contains 1 unit kg of vitamin A}
\displaystyle \text{and 2 units kg of vitamin C. It costs Rs.5.00 per kg to purchase food 'I' and Rs.7.00 per kg}
\displaystyle \text{to produce food 'II'. Determine the minimum cost to such a mixture. Formulate the above as a}
\displaystyle \text{LPP and solve it. [CBSE 2012]}

\displaystyle \text{Answer:}

\displaystyle  \text{Let the dietician mix }x\text{ kg of food 'I' with }y\text{ kg of food 'II'. Then,}
\displaystyle \text{the mathematical model of the LPP is as follows:}
\displaystyle \text{Minimize }Z=5x+7y
\displaystyle \text{Subject to:} \\ 2x+y \ge 8
\displaystyle x+2y \ge 10
\displaystyle x,\ y \ge 0
\displaystyle \text{Thus, the shaded region is the feasible region of the LPP.}\displaystyle \text{The coordinates of the corner-points of this region are }(10,0),\ (2,4)\text{ and }(0,8).
\displaystyle \text{The point }(2,4)\text{ is obtained by solving }2x+y=8\text{ and }x+2y=10\text{ simultaneously.}
\displaystyle \text{The values of the objective function }Z=5x+7y\text{ at the corner points of the feasible}
\displaystyle \text{region are given in the following table:}
\displaystyle \begin{array}{c|c} \text{Point }(x,y) & \text{Value of the objective function }Z=5x+7y\\ \hline (10,0) & Z=5\times10+7\times0=50\\ (2,4) & Z=5\times2+7\times4=38\\ (0,8) & Z=5\times0+7\times8=56 \end{array}
\displaystyle \text{Clearly, }Z\text{ is minimum at }x=2\text{ and }y=4.\ \text{The minimum value of }Z\text{ is }38.
\displaystyle \text{We observe that open half plane represented by }5x+7y<38\text{ does not have points in common}
\displaystyle \text{with the feasible region. So, }Z\text{ has minimum value equal to }38\text{ at }x=2\text{ and }y=4.
\displaystyle \text{Hence, the optimal mixing strategy for the dietician will be to mix }2\text{ kg of food 'I'}
\displaystyle \text{and }4\text{ kg of food 'II'. In this case, his cost will be minimum and the minimum}
\displaystyle \text{cost will be Rs.38.00.}

\displaystyle \textbf{Question 4:} \\  \text{A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work}
\displaystyle \text{on machine A and 3 hours on machin B to produce a package of nuts while it takes 3 hours on}
\displaystyle \text{machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of}
\displaystyle \text{Rs.2.50 per package of nuts and Re.1.00 per package of bolts. How many packages of each}
\displaystyle \text{should he produce each day so as to maximize his profit, if he operates his machines for at}
\displaystyle \text{most 12 hours a day? Formulate this mathematically and then solve it. [CBSE 2012]}

\displaystyle \text{Answer:}
\displaystyle \text{The given information can be summarized in the following tabular form:}
\displaystyle \begin{array}{c|cc|c} \text{Machines} & \multicolumn{2}{c|}{\text{Time required to produce products}} & \text{Max. Machine hours available}\\ & \text{Nut} & \text{Bolt} & \\ \hline A & 1 & 3 & 12\\ B & 3 & 1 & 12\\ \hline \text{Profit (in Rs)} & 2.50 & 1.00 & \end{array}
\displaystyle \text{Let the manufacturer produce }x\text{ packages of nuts and }y\text{ packages of bolts each day.}
\displaystyle \text{Since machine A takes one hour to produce one package of nuts and 3 hours to produce one}
\displaystyle \text{package of bolts. Therefore, the total time required by machine A to produce }x\text{ packages of nuts}
\displaystyle \text{and }y\text{ packages of bolts is }(x+3y)\text{ hours. But machine A operates for at most 12 hours.}
\displaystyle x+3y \le 12
\displaystyle \text{Similarly, the total time required by machine B to produce }x\text{ packages of nuts and }y\text{ packages of}
\displaystyle \text{bolts is }(3x+y)\text{ hours. But machine B operates for at most 12 hours.}
\displaystyle 3x+y \le 12
\displaystyle \text{Since the profit on one package of nuts is Rs.2.50 and on one package of bolts the profit is Rs.1.}
\displaystyle \text{Therefore, profit on }x\text{ packages of nuts and }y\text{ packages of bolts is Rs.(2.50x+y). Let }Z\text{ denote}
\displaystyle \text{the total profit. Then, }Z=2.50x+y.
\displaystyle \text{Clearly, }x \ge 0\ \text{and}\ y \ge 0
\displaystyle \text{Thus, the above LPP can be stated mathematically as follows:}
\displaystyle \text{Maximize }Z=2.50x+y
\displaystyle \text{Subject to:} \\ x+3y \le 12
\displaystyle 3x+y \le 12
\displaystyle \text{and, }x,\ y \ge 0
\displaystyle \text{To solve this LPP graphically, we first convert the inequations into equations to obtain the}
\displaystyle \text{following equations.}
\displaystyle x+3y=12,\ 3x+y=12,\ x=0,\ y=0
\displaystyle \text{Thus, the shaded region represents the feasible region of the given LPP.}\displaystyle \text{The coordinates of the corner-points of the feasible region }\text{ are }(0,0),(4,0),
\displaystyle (3,3)\ \text{and}\ (0,4).\ \text{These points are obtained by solving the corresponding intersecting lines}
\displaystyle \text{simultaneously.}
\displaystyle \text{The values of the objective function at the corner-points of the feasible region are given in the}
\displaystyle \text{following table:}
\displaystyle \begin{array}{c|c} \text{Point }(x,y) & \text{Value of the objective function }Z=2.50x+y\\ \hline (0,0) & Z=2.50\times0+1\times0=0\\ (4,0) & Z=2.50\times4+1\times0=10\\ (3,3) & Z=2.50\times3+1\times3=10.50\\ (0,4) & Z=2.50\times0+1\times4=4 \end{array}
\displaystyle \text{Clearly, }Z\text{ is maximum at }x=3,\ y=3\text{ and the maximum value of }Z\text{ is }10.50.
\displaystyle \text{Hence, the optimal production strategy for the manufacturer will be to manufacture 3 packages}
\displaystyle \text{each of nuts and bolts daily and in this case his maximum profit will be Rs.10.50.}

\displaystyle \textbf{Question 5:} \\ \text{An oil company requires 12,000, 20,000 and 15,000 barrels of high-grade, medium grade}
\displaystyle \text{and low grade oil, respectively. Refinery A produces 100, 300 and 200 barrels per day}
\displaystyle \text{of high-grade, medium-grade and low-grade oil, respectively, while refinery B produces}
\displaystyle \text{200, 400 and 100 barrels per day of high-grade, medium-grade and low-grade oil.}
\displaystyle \text{If refinery A costs Rs.400 per day and refinery B costs Rs.300 per day to operate,}
\displaystyle \text{how many days should each be run to minimize costs while satisfying requirements.}
\displaystyle \text{[CBSE 2004]}

\displaystyle \text{Answer:}
\displaystyle  \text{The given data may be put in the following tabular form:}
\displaystyle \begin{array}{c|ccc|c} \text{Refinery} & \text{High-grade} & \text{Medium-grade} & \text{Low-grade} & \text{Cost per day}\\ \hline A & 100 & 300 & 200 & \text{Rs.400}\\ B & 200 & 400 & 100 & \text{Rs.300}\\ \hline \text{Minimum Requirement} & 12,000 & 20,000 & 15,000 & \end{array}
\displaystyle \text{Suppose refineries A and B should run for }x\text{ and }y\text{ days respectively to minimize}
\displaystyle \text{the total cost. The mathematical form of the above LPP is}
\displaystyle \text{Minimize}\ Z=400x+300y
\displaystyle \text{Subject to}
\displaystyle 100x+200y \ge 12,000
\displaystyle 300x+400y \ge 20,000
\displaystyle 200x+100y \ge 15,000
\displaystyle \text{and,}\ x,\ y \ge 0
\displaystyle \text{The feasible region of the above LPP is represented by the shaded region below:}\displaystyle \text{The corner points of the feasible region are }(120,0),\ (60,30)\text{ and }(0,150).
\displaystyle \text{The value of the objective function at these points are given in the following table:}
\displaystyle \begin{array}{c|c} \text{Point }(x,y) & \text{Value of the objective function }Z=400x+300y\\ \hline (120,0) & Z=400\times120+300\times0=48000\\ (60,30) & Z=400\times60+300\times30=33000\\ (0,150) & Z=400\times0+300\times150=45000 \end{array}
\displaystyle \text{Clearly, }Z\text{ is minimum when }x=60,\ y=30.\ \text{The feasible region is unbounded.}
\displaystyle \text{So, we find the half-plane represented by }400x+300y<33000.
\displaystyle \text{Clearly, the half-plane does not have points in common with the feasible region.}
\displaystyle \text{So, }Z\text{ is minimum at }x=60,\ y=30.
\displaystyle \text{Hence, the machine A should run for 60 days and the machine B should run for 30 days to}
\displaystyle \text{minimize the cost while satisfying the constraints.}

\displaystyle \textbf{Question 6:} \\ \text{A company produces soft drinks that has a contract which requires that a}
\displaystyle \text{minimum of 80 units of the chemical A and 60 units of the chemical B to go}
\displaystyle \text{into each bottle of the drink. The chemicals are available in a prepared mix}
\displaystyle \text{from two different suppliers. Supplier S has a mix of 4 units of A and 2}
\displaystyle \text{units of B that costs Rs.10, the supplier T has a mix of 1 unit of A and 1 unit}
\displaystyle \text{of B that costs Rs.4. How many mixes from S and T should the company purchase}
\displaystyle \text{to honour contract requirement and yet minimize cost?}
\displaystyle \text{[CBSE 2012]}

\displaystyle \text{Answer:}
\displaystyle  \text{The given data may be put in the following tabular form:}
\displaystyle \begin{array}{c|cc|c} \text{Chemical} & \text{S} & \text{T} & \text{Minimum Requirement}\\ \hline A & 4 & 1 & 80\\ B & 2 & 1 & 60\\ \hline \text{Cost per unit} & \text{Rs.10} & \text{Rs.4} & \end{array}
\displaystyle \text{Suppose }x\text{ units of mix are purchased from supplier S and }y\text{ units are purchased }
\displaystyle \text{from supplier T. Total cost }Z=10x+4y.
\displaystyle \text{Units of chemical A per bottle }=4x+y.\ \text{But the minimum requirement of}
\displaystyle \text{chemical A per bottle is 80 units.}
\displaystyle 4x+y \ge 80
\displaystyle \text{Similarly, }2x+y \ge 60.\ \text{Clearly, }x \ge 0,\ y \ge 0
\displaystyle \text{Thus, the mathematical formulation of the given LPP is}
\displaystyle \text{Minimize}\ Z=10x+4y
\displaystyle \text{Subject to:}
\displaystyle 4x+y \ge 80
\displaystyle 2x+y \ge 60
\displaystyle \text{and,}\ x \ge 0,\ y \ge 0
\displaystyle \text{Now, we find the feasible region which is the set of all points whose coordinates}\displaystyle \text{The shaded region in Figure represents the feasible region of the given LPP.}
\displaystyle \text{The coordinates of the corner points of the feasible region are }(30,0),
\displaystyle (10,40),\ (0,80).
\displaystyle \text{These points are obtained by solving the equations of the corresponding intersecting}
\displaystyle \text{lines, simultaneously. The values of the objective function at these points are}
\displaystyle \text{given in the following table:}
\displaystyle \begin{array}{c|c} \text{Point }(x,y) & \text{Value of objective function }Z=10x+4y\\ \hline (30,0) & Z=10\times30+4\times0=300\\ (10,40) & Z=10\times10+40\times4=260\\ (0,80) & Z=10\times0+4\times80=320 \end{array}
\displaystyle \text{Clearly, }Z\text{ is minimum at }(10,40). \text{The feasible region is unbounded and}
\displaystyle \text{the open half plane represented by }10x+4y<260\text{ does not have points in common}
\displaystyle \text{with the feasible region. So, }Z\text{ is minimum at }x=10,\ y=40.
\displaystyle \text{Hence, the cost per bottle is minimum when the company purchases 10 mixes from supplier S}
\displaystyle \text{and 40 mixes from supplier T.}

\displaystyle \textbf{Question 7:} \\  \text{A dealer wishes to purchase a number of fans and sewing machines.}
\displaystyle \text{He has only Rs.5760.00 to invest and has space for at most 20 items. A fan costs}
\displaystyle \text{him Rs.360.00 and a sewing machine Rs.240.00. His expectation is that he can sell}
\displaystyle \text{a fan at a profit of Rs.22.00 and a sewing machine at a profit of Rs.18.00.}
\displaystyle \text{Assuming that he can sell all the items that he can buy, how should he invest his}
\displaystyle \text{money in order to maximize his profit? Translate this problem mathematically and}
\displaystyle \text{then solve it. [CBSE 2001C, 2002C]}

\displaystyle \text{Answer:}
\displaystyle  \text{Suppose the dealer buys }x\text{ fans and }y\text{ sewing machines. Since the dealer has}
\displaystyle \text{space for at most 20 items. Therefore,}
\displaystyle x+y \le 20
\displaystyle \text{A fan costs Rs.360 and a sewing machine costs Rs.240. Therefore, total cost of }x\text{ fans and }y
\displaystyle \text{sewing machines is Rs.(360x+240y). But the dealer has only Rs.5760 to invest. Therefore,}
\displaystyle 360x+240y \le 5760
\displaystyle \text{Since the dealer can sell all the items that he can buy and the profit on a fan is Rs.22 and on a}
\displaystyle \text{sewing machine the profit is Rs.18. Therefore, total profit on selling }x\text{ fans and }y\text{ sewing machines is}
\displaystyle \text{Rs.(22x+18y). Let }Z\text{ denote the total profit. Then, }Z=22x+18y.
\displaystyle \text{Clearly, }x,\ y \ge 0.
\displaystyle \text{Thus, the mathematical formulation of the given problem is}
\displaystyle \text{Maximize}\ Z=22x+18y
\displaystyle \text{Subject to}
\displaystyle x+y \le 20
\displaystyle 360x+240y \le 5760
\displaystyle \text{and,}\ x \ge 0,\ y \ge 0
\displaystyle \text{The feasible region of the LPP is shaded in Figure below}\displaystyle \text{The feasible region  points }\text{ are }(0,0),\ (16,0),\ (8,12)\text{ and }(0,20).
\displaystyle \text{These points have been obtained by solving the corresponding intersecting lines, simultaneously.}
\displaystyle \text{The values of the objective function }Z\text{ at corner-points of the feasible region are given in the}
\displaystyle \text{following table.}
\displaystyle \begin{array}{c|c} \text{Point }(x,y) & \text{Value of the objective function }Z=22x+18y\\ \hline (0,0) & Z=22\times0+18\times0=0\\ (16,0) & Z=22\times16+18\times0=352\\ (8,12) & Z=22\times8+18\times12=392\\ (0,20) & Z=22\times0+20\times18=360 \end{array}
\displaystyle \text{Clearly, }Z\text{ is maximum at }x=8\text{ and }y=12.\ \text{The maximum value of }Z\text{ is }392.
\displaystyle \text{Hence, the dealer should purchase 8 fans and 12 sewing machines to obtain the maximum profit}
\displaystyle \text{under given conditions.}

\displaystyle \textbf{Question 8. }\text{The corner points of the feasible region in the graphical}
\displaystyle \text{representation of a linear programming problem are }(2,72),\ (15,20)\text{ and }(40,15).
\displaystyle \text{If }Z=18x+9y\text{ be the}  \text{ objective function, then} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }Z\text{ is maximum at }(2,72),\text{ minimum at }(15,20).
\displaystyle \text{(b) }Z\text{ is maximum at }(15,20),\text{ minimum at }(40,15).
\displaystyle \text{(c) }Z\text{ is maximum at }(40,15),\text{ minimum at }(15,20).
\displaystyle \text{(d) }Z\text{ is maximum at }(40,15),\text{ minimum at }(2,72).
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }Z=18x+9y
\displaystyle \text{At }(2,72),\ Z=18\times2+9\times72=684
\displaystyle \text{At }(15,20),\ Z=18\times15+9\times20=450
\displaystyle \text{At }(40,15),\ Z=18\times40+9\times15=855
\displaystyle \text{Thus, }Z\text{ is maximum at }(40,15)\text{ and minimum at }(15,20).
\\

\displaystyle \textbf{Question 9. }\text{The number of corner points of the feasible region determined by}
\displaystyle \text{the constraints } x-y\geq0,\ 2y\leq x+2,\ x\geq0\text{ and }y\geq0\text{ is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }2 \qquad \text{(b) }3
\displaystyle \text{(c) }4 \qquad \text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, subject to constraints}
\displaystyle x-y\geq0,\ 2y\leq x+2,\text{ and }x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle x-y=0\qquad \cdots(i)
\displaystyle \text{and }2y=x+2
\displaystyle 2y-x=2\qquad \cdots(ii)
\displaystyle \text{Table for line }x-y=0\text{ is}
\displaystyle \begin{array}{c|c}x&0\\ \hline y&0\end{array}
\displaystyle \text{Table for line }2y-x=2\text{ is}
\displaystyle \begin{array}{c|cc}x&0&-2\\ \hline y&1&0\end{array}

\displaystyle \therefore \text{The corner points are }O(0,0)\text{ and }A(2,2).
\\

\displaystyle \textbf{Question 10. }\text{The objective function }Z=ax+by\text{ of an LPP if its maximum value }42
\displaystyle \text{at }(4,6)\text{ and minimum value }19\text{ at }(3,2).\text{ Which of the following is true?} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }a=9\text{ and }b=1 \qquad \text{(b) }a=5\text{ and }b=2
\displaystyle \text{(c) }a=3\text{ and }b=5 \qquad \text{(d) }a=5\text{ and }b=3
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }Z=ax+by
\displaystyle \text{At }(4,6),\ Z=ax+by
\displaystyle 42=4a+b\times6
\displaystyle 42=4a+6b
\displaystyle 21=2a+3b\qquad \cdots(i)
\displaystyle \text{At }(3,2),\ 19=3a+2b\qquad \cdots(ii)
\displaystyle \text{On multiplying Eq. (i) by }2\text{ and Eq. (ii) by }3\text{ and solving, we get}
\displaystyle 4a+6b=42
\displaystyle 9a+6b=57
\displaystyle \therefore -5a=-15
\displaystyle \Rightarrow a=3
\displaystyle \text{Substitute }a=3,\text{ in Eq. (i), we get}
\displaystyle 2\times3+3b=21
\displaystyle \Rightarrow b=5
\\

\displaystyle \textbf{Question 11. }\text{The corner points of the feasible region of linear programming problem are}
\displaystyle (0,4),\ (8,0)\text{ and }\left(\frac{20}{4},\frac{4}{3}\right).\text{ If }Z=30x+24y\text{ is the objective function, then}
\displaystyle (\text{Maximum value of }Z-\text{Minimum value of }Z)\text{ is equal to} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }144 \qquad \text{(b) }96
\displaystyle \text{(c) }120 \qquad \text{(d) }136
\displaystyle \text{Answer:}
\displaystyle \text{(a) }Z=30x+24y\qquad \text{(given)}
\displaystyle \text{At }(0,4),\ Z=30\times0+24\times4=96\text{ (minimum)}
\displaystyle \text{At }(8,0),\ Z=30\times8+24\times0=240\text{ (maximum)}
\displaystyle \text{At }\left(\frac{20}{3},\frac{4}{3}\right),\ Z=30\times\frac{20}{3}+24\times\frac{4}{3}=232
\displaystyle \text{Now, (Maximum value of }Z\text{ - Minimum value of }Z)
\displaystyle =240-96=144
\\

\displaystyle \textbf{Question 12. }\text{The feasible region of a linear programming problem is shown in} \\ \text{the figure below:} \hspace{0.2cm}\text{[CBSE 2023]} \displaystyle \text{Which of the following are the possible constraints?}
\displaystyle \text{(a) }x+2y\geq4,\ x+y\leq3,\ x\geq0,\ y\geq0
\displaystyle \text{(b) }x+2y\leq4,\ x+y\leq3,\ x\geq0,\ y\geq0
\displaystyle \text{(c) }x+2y\geq4,\ x+y\geq3,\ x\geq0,\ y\geq0
\displaystyle \text{(d) }x+2y\geq4,\ x+y\geq3,\ x\leq0,\ y\leq0
\displaystyle \text{Answer:}
\displaystyle \text{(c) If we put }O(0,0)\text{ in }x+2y\geq4\text{ inequality, then it will be false and it will }
\displaystyle \text{be away from origin. So, }x+2y\geq4\text{ is possible.}
\displaystyle \text{Similarly, the inequality }x+y\geq3\text{ will also be away from origin, which means}
\displaystyle \text{ it is false when }O(0,0)\text{ is put in inequality.}
\displaystyle \text{So, }x+y\geq3\text{ is possible.}
\displaystyle \therefore \text{It is clear that feasible region lies in Ist quadrant, so }x\geq0,\ y\geq0
\\

\displaystyle \textbf{Question 13. }\text{Solve the following linear programming problem graphically:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Minimize }Z=x+2y
\displaystyle \text{Subject to the constraints: }2x+y\geq3,\ x+2y\geq6,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }Z=x+2y
\displaystyle \text{Subject to the constraints}
\displaystyle 2x+y\geq3;\ x+2y\geq6\text{ and }x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle 2x+y=3\qquad \cdots(i)
\displaystyle x+2y=6\qquad \cdots(ii)
\displaystyle \text{Table for line }2x+y=3\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&1.5&0\\ \hline y&0&3\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+y\geq3  0\geq3\text{ (which is false)}
\displaystyle \text{So, the half plane, is away from the origin.}
\displaystyle \text{Table for line }x+2y=6
\displaystyle \begin{array}{|c|cc|} \hline x&6&0\\ \hline y&0&3\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\geq6,\ 0\geq6\text{ (which is false).}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ so the feasible region lies in the Ist quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }(0,3).
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \therefore \text{Feasible region is shaded region above.}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=x+2y\\ \hline A(0,3) & 6\ \text{(Minimum)}\\ \hline B(6,0) & 6\ \text{(Minimum)}\\ \hline \end{array}
\displaystyle \text{Here, the feasible region is unbounded and the open half plane determined by}
\displaystyle x+2y<6\text{ has no point in common with the feasible region.} \\ \\ \text{Hence, }Z=6\text{ is minimum }  \text{at }A(0,3)\text{ and }C(6,0)\text{ and also on the line segment }AC.
\\

\displaystyle \textbf{Question 14. }\text{Solve the following linear programming problem graphically:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Minimize }Z=-3x+4y
\displaystyle \text{Subject to constraints: }x+2y\leq8,\ 3x+2y\leq12,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }Z=-3x+4y
\displaystyle \text{Subject to the constraints}
\displaystyle x+2y\leq8;\ 3x+2y\leq12\text{ and }x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle x+2y=8\qquad \cdots(i)
\displaystyle 3x+2y=12\qquad \cdots(ii)
\displaystyle \text{Table for line }x+2y=8\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&8&0\\ \hline y&0&4\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq8  0\leq8\text{ (which is true)}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Table for line }3x+2y=12
\displaystyle \begin{array}{|c|cc|} \hline x&4&0\\ \hline y&0&6\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+2y\leq12,\ 0\leq12\text{ (which is true).}
\displaystyle \text{So, half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0\text{ and }y\geq0,\text{ so the feasible region lies in the Ist quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }(2,3).
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=-3x+4y\\ \hline A(0,4) & 16\\ \hline B(2,3) & 6\\ \hline C(4,0) & -12\ \text{(Minimum)}\\ \hline O(0,0) & 0\\ \hline \end{array}
\displaystyle \text{Hence, }Z=-12\text{ is minimum at }(4,0).
\\

\displaystyle \textbf{Question 15. }\text{Solve the following linear programming problem graphically:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Maximize }Z=-3x-5y
\displaystyle \text{Subject to the constraints: }-2x+y\leq4,\ x+y\geq3,\ x-2y\leq2,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Maximize }Z=-3x-5y
\displaystyle \text{Subject to the constraints}
\displaystyle -2x+y\leq4,
\displaystyle x+y\geq3,
\displaystyle x-2y\leq2
\displaystyle \text{and }x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle -2x+y=4\qquad \cdots(i)
\displaystyle x+y=3\qquad \cdots(ii)
\displaystyle x-2y=2\qquad \cdots(iii)
\displaystyle \text{Table for line }-2x+y=4\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&-2&0\\ \hline y&0&4\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(-2,0)\text{ and }(0,4)
\displaystyle \text{On putting }(0,0)\text{ in the inequality }-2x+y\leq4,  \text{ we have }  0\leq4\text{ (which is true)}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }x+y=3\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&3&0\\ \hline y&0&3\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(3,0)\text{ and }(0,3).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\geq3,  0+0\geq3\text{ (which is false)}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for line }x-2y=2\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&2&0\\ \hline y&0&-1\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(2,0)\text{ and }(0,-1).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x-2y\leq2,\ 0\leq2\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 Ist quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }\left(-\frac{1}{3},\frac{10}{3}\right),
\displaystyle \text{Eqs. (ii) and (iii) is }\left(\frac{8}{3},\frac{1}{3}\right)\text{ and Eqs. (i) and (iii) is } \left(-\frac{10}{3},-\frac{8}{3}\right).
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \therefore \text{Clearly, feasible region is shaded.}
\displaystyle \text{The value of }Z\text{ at corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=-3x-5y\\ \hline A\left(\frac{8}{3},\frac{1}{3}\right) & -3\times\frac{8}{3}-5\times\frac{1}{3}=-8-\frac{5}{3}=-\frac{29}{3}\ \text{(Maximum)}\\ \hline B(0,3) & -3\times0-5\times3=-15\\ \hline C(0,4) & -3\times0-5\times4=-20\\ \hline \end{array}
\displaystyle \text{Here, the feasible region is unbounded and the open half plane determined by}
\displaystyle -3x-5y>-\frac{29}{3}\text{ has no point } \text{in common with the feasible region. Hence, }
\displaystyle Z=-\frac{29}{3}\text{ is}  \text{ maximum at }\left(\frac{8}{3},\frac{1}{3}\right).
\\

\displaystyle \textbf{Question 16. }\text{Solve graphically the following linear programming problem:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Maximize }Z=6x+3y
\displaystyle \text{Subject to the constraints: }4x+y\geq80,\ 3x+2y\leq150,\ x+5y\geq115, \\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{We have the following LPP,}
\displaystyle \text{Maximize }Z=6x+3y
\displaystyle \text{Subject to the constraints}
\displaystyle 4x+y\geq80,
\displaystyle 3x+2y\leq150,
\displaystyle x+5y\geq115
\displaystyle \text{and }x\geq0,\ y\geq0.
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle 4x+y=80\qquad \cdots(i)
\displaystyle 3x+2y=150\qquad \cdots(ii)
\displaystyle x+5y=115\qquad \cdots(iii)
\displaystyle \text{Table for line }4x+y=80\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&20&0\\ \hline y&0&80\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(20,0)\text{ and }(0,80).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }4x+y\geq80 \  0\geq80\text{ (which is false).}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for line }3x+2y=150\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&50&0\\ \hline y&0&75\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(50,0)\text{ and }(0,75).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+2y\leq150 \  0\leq150\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }x+5y=115\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&115&0\\ \hline y&0&23\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(115,0)\text{ and }(0,23).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+5y\geq115,\ 0\geq115\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 Ist quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }(2,72),\text{ Eqs. (ii) and (iii) is }(40,15)
\displaystyle \text{and Eqs. (i) and (iii) is }(15,20).
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \text{Clearly, feasible region is }PQR,\text{ whose corner points are }P(2,72),\ Q(15,20),\ R(40,15).
\displaystyle \text{The value of }Z\text{ at corner points are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=6x+3y\\ \hline P(2,72) & 6\times2+3\times72=228\\ \hline Q(15,20) & 6\times15+3\times20=150\\ \hline R(40,15) & 6\times40+3\times15=285\ \text{(Maximum)}\\ \hline \end{array}
\displaystyle \text{In the table, we find that maximum value of }Z\text{ is }285  \text{ at }R(40,15).
\\

\displaystyle \textbf{Question 17. }\text{Solve the following LPP graphically:} \hspace{0.2cm}\text{[CBSE 2023; CBSE 2017]}
\displaystyle \text{Minimise }Z=5x+10y
\displaystyle \text{Subject to the constraints }x+2y\leq120,\ x+y\geq60,\ x-2y\geq0,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Our problem is to minimise}
\displaystyle Z=5x+10y\qquad \cdots(i)
\displaystyle \text{Subject to constraints}
\displaystyle x+2y\leq120\qquad \cdots(ii)
\displaystyle x+y\geq60\qquad \cdots(iii)
\displaystyle x-2y\geq0\qquad \cdots(iv)
\displaystyle \text{and}
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{Table for line }x+2y=120\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&120\\ \hline y&60&0\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in inequality }x+2y\leq120,\text{ we get }0+2\times0\leq120
\displaystyle \Rightarrow 0\leq120\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin. Secondly, table for the line } \\ x+y=60\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&60\\ \hline y&60&0\\ \hline \end{array}
\displaystyle \text{On putting }(0,0)\text{ in inequality }x+y\geq60,\text{ we get }0+0\geq60
\displaystyle \Rightarrow 0\geq60\text{ (which is false).}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for the line }x-2y=0\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&10\\ \hline y&0&5\\ \hline \end{array}
\displaystyle \text{On putting }(5,0)\text{ in the inequality }x-2y\geq0,\text{ we get }5-2\times0\geq0
\displaystyle \Rightarrow 5\geq0\text{ (which is true).}
\displaystyle \text{Thus, the half plane is towards the X-axis. Since, }x,y\geq0
\displaystyle \therefore \text{The feasible region lies in the Ist quadrant.}\displaystyle \text{Clearly, feasible region is }ABCDA.
\displaystyle \text{On solving equations }x-2y=0\text{ and }x+y=60,\text{ we get }D(40,20)\text{ and on solving}
\displaystyle \text{equations }x-2y=0\text{ and }x+2y=120,\text{ we get }C(60,30).
\displaystyle \text{The corner points of the feasible region are }A(60,0),\ B(120,0),\ C(60,30)
\displaystyle \text{and }D(40,20).\text{ The values of }Z\text{ at these points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=5x+10y\\ \hline A(60,0) & 300\ \text{(Minimum)}\\ \hline B(120,0) & 600\\ \hline C(60,30) & 600\\ \hline D(40,20) & 400\\ \hline \end{array}
\displaystyle \text{Clearly, the minimum value of }Z\text{ is }300\text{ at the point }A(60,0).
\\

\displaystyle \textbf{Question 18. }\text{Solve the following LPP graphically:} \hspace{0.2cm}\text{[CBSE 2023; CBSE 2017]}
\displaystyle \text{Maximise and minimise }Z=x+2y
\displaystyle \text{Subject to the constraints }x+2y\geq100,\ 2x-y\leq0,\ 2x+y\leq200, \\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Our problem is to minimise and maximise}
\displaystyle Z=x+2y\qquad \cdots(i)
\displaystyle \text{Subject to constraints}
\displaystyle x+2y\geq100\qquad \cdots(ii)
\displaystyle 2x-y\leq0\qquad \cdots(iii)
\displaystyle 2x+y\leq200\qquad \cdots(iv)
\displaystyle \text{and}
\displaystyle x\geq0,\ y\geq0\qquad \cdots(v)
\displaystyle \text{Table for line }x+2y=100\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&100\\ \hline y&50&0\\ \hline \end{array}
\displaystyle \text{So, the line }x+2y=100\text{ is passing through the points }  (0,50)\text{ and }(100,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\geq100,\text{ we get }  0+2\times0\geq100
\displaystyle \Rightarrow 0\geq100\text{ (which is false)}
\displaystyle \text{So, the half plane is away from the origin.}
\displaystyle \text{Table for line }2x+y=200\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&100\\ \hline y&200&0\\ \hline \end{array}
\displaystyle \text{So, the line }2x+y=200\text{ is passing through the points }(0,200)\text{ and }(100,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+y\leq200,\text{ we get }2\times0+0\leq200
\displaystyle \Rightarrow 0\leq200\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x,y\geq0.
\displaystyle \text{So, the region lies in the Ist quadrant.}\displaystyle \text{Clearly, feasible region is }ABCDA.
\displaystyle \text{On solving equations }2x-y=0\text{ and }x+2y=100,\text{ we get }B(20,40).
\displaystyle \text{The maximum value of }Z\text{ is }400\text{ at }D(0,200)\text{ and the minimum value of }Z
\displaystyle \text{is }100\text{ at all the points on the line segment joining }A(0,50)\text{ and }B(20,40).
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\displaystyle \textbf{Question 19. }\text{The graph of the inequality }2x+3y>6\text{ is} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) half plane that contains the origin}
\displaystyle \text{(b) half plane that neither contains the origin nor the points of the line } \\ 2x+3y=6
\displaystyle \text{(c) whole }XOY\text{-plane excluding the points on the line }2x+3y=6
\displaystyle \text{(d) entire }XOY\text{-plane}
\displaystyle \text{Answer:}
\displaystyle \text{(b) The inequality }2x+3y>6\text{ represents half plane that neither contains}
\displaystyle \text{the origin nor the points of the line }2x+3y=6.
\\

\displaystyle \textbf{Question 20. }\text{Solve the following Linear Programming Problem graphically:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Maximize }Z=70x+40y
\displaystyle \text{Subject to constraints }3x+2y\leq9,\ 3x+y\leq9,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{Maximise }Z=70x+40y
\displaystyle \text{Subject to the constraints}
\displaystyle 3x+2y\leq9,
\displaystyle 3x+y\leq9
\displaystyle \text{and}
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle 3x+2y=9\qquad \cdots(i)
\displaystyle 3x+y=9\qquad \cdots(ii)
\displaystyle \text{Table for line }3x+2y=9\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&3&0\\ \hline y&0&9/2\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(3,0)\text{ and }(0,9/2).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+2y\leq9 \  0\leq9\text{ (which is true)}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }3x+y=9\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&3&0\\ \hline y&0&9\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(3,0)\text{ and }(0,9).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+y\leq9,\ 0\leq9\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 Ist quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }(3,0).
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \text{Clearly, feasible region is }OABO,\text{ whose corner points are }O(0,0),
\displaystyle A(3,0)\text{ and }B(0,9/2).
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=70x+40y\\ \hline O(0,0) & 0\\ \hline A(3,0) & 210\ \text{(Maximum)}\\ \hline B(0,9/2) & 180\\ \hline \end{array}
\displaystyle \text{In the table, we find that maximum value of }Z\text{ is }210  \text{ at the point }A(3,0).
\\

\displaystyle \textbf{Question 21. }\text{Find the minimum and maximum values of the objective function}  \\  Z=3x+9y   \\ \text{Subject to constraints } \\ x+3y\leq60,\ x+y\geq10,\ x\leq y,\ x\geq0,  y\geq0. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given that,}
\displaystyle \text{Minimise and Maximise }Z=3x+9y\qquad \cdots(i)
\displaystyle \text{Subject to constraints are}
\displaystyle x+3y\leq60\qquad \cdots(ii)
\displaystyle x+y\geq10\qquad \cdots(iii)
\displaystyle x\leq y\qquad \cdots(iv)
\displaystyle \text{and}
\displaystyle x\geq0,\ y\geq0\qquad \cdots(v)
\displaystyle \text{First of all, let us plot the graph of the feasible region of the system of linear}
\displaystyle \text{inequalities (ii) to (v). The feasible region }ABCDA\text{ is shown in the figure.}\displaystyle \text{Note that the region is bounded. The coordinates of the corner points }A,\ B,\ C\text{ and }D
\displaystyle \text{are }(0,10),\ (5,5),\ (15,15)\text{ and }(0,20),\text{ respectively.}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Corresponding value of }Z=3x+9y\\ \hline A(0,10) & 90\\ \hline B(5,5) & 60\ \text{(Minimum)}\\ \hline C(15,15) & 180\ \text{(Maximum)}\\ \hline D(0,20) & 180\ \text{(Multiple optimal solutions)}\\ \hline \end{array}
\displaystyle \text{We, now find the minimum and maximum value of }Z.
\displaystyle \text{From the table, we find that the minimum value of }Z\text{ is }60\text{ at the point }B(5,5)
\displaystyle \text{of the feasible region.}
\displaystyle \text{The maximum value of }Z\text{ on the feasible region occurs at the two corner points }C(15,15)
\displaystyle \text{and }D(0,20)\text{ and it is }180\text{ in each case.}
\displaystyle \text{Remark: Observe that in the above example, the problem has multiple optimal solutions}
\displaystyle \text{at the corner points }C\text{ and }D,\text{ i.e. both points produce same maximum value }180.
\displaystyle \text{In such cases, you can see that every point on the line segment }CD\text{ joining the two}
\displaystyle \text{corner points }C\text{ and }D\text{ also gives same maximum value. Same is also true in the case,}
\displaystyle \text{if the two points produce same minimum value.}
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\displaystyle \textbf{Question 22. }\text{Solve the following linear programming problem graphically:} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Maximize }Z=34x+45y
\displaystyle \text{Subject to constraints }x+y\leq300,\ 2x+3y\leq70,\ x\geq0,\ y\geq0.
\displaystyle \text{Answer:}
\displaystyle \text{We have the following LPP}
\displaystyle \text{Maximize }Z=34x+45y
\displaystyle \text{Subject to the constraints}
\displaystyle x+y\leq300,\ 2x+3y\leq70\text{ and }x,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle x+y=300\qquad \cdots(i)
\displaystyle \text{and}
\displaystyle 2x+3y=70\qquad \cdots(ii)
\displaystyle \text{Table for line }x+y=300\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&0&300\\ \hline y&300&0\\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(0,300)\text{ and }(300,0).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq300,\text{ we get }  0+0\leq300\text{ (which is true)}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }2x+3y=70\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&35&0\\ \hline y&0&70/3\\ \hline \end{array}
\displaystyle \text{So, the line passes through the points }(35,0)\text{ and }(0,70/3).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+3y\leq70,\text{ we get }0+0\leq70
\displaystyle \text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x,\ y\geq0,\text{ so the region lies in the Ist quadrant.}
\displaystyle \text{The graphical representation of the above system of inequations is given below.}\displaystyle \text{Clearly, feasible region is }OABO,\text{ where the corner points are }O(0,0),
\displaystyle A(35,0)\text{ and }B\left(0,\frac{70}{3}\right).
\displaystyle \text{Now, the values of }Z\text{ at corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=34x+45y\\ \hline O(0,0) & 0+0=0\\ \hline A(35,0) & 34\times35+45\times0=1190\ \text{(Maximum)}\\ \hline B\left(0,\frac{70}{3}\right) & 34\times0+45\times\frac{70}{3}=1050\\ \hline \end{array}
\displaystyle \text{Hence, the maximum value of }Z\text{ is }1190\text{ at }(35,0).
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\displaystyle \textbf{Question 23. }\text{Find graphically, the maximum value of } \\ Z=2x+5y.  \\  \text{Subject to constraints } \\ 2x+4y\leq8,\ 3x+y\leq6,\ x+y\leq4,\ x\geq0,  y\geq0. \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have the following LPP,}
\displaystyle \text{Maximize }Z=2x+5y
\displaystyle \text{Subject to the constraints}
\displaystyle 2x+4y\leq8\text{ or }x+2y\leq4,\quad 3x+y\leq6,\quad x+y\leq4
\displaystyle \text{and}
\displaystyle x\geq0,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle x+2y=4\qquad \cdots(i)
\displaystyle 3x+y=6\qquad \cdots(ii)
\displaystyle \text{and}
\displaystyle x+y=4\qquad \cdots(iii)
\displaystyle \text{Table for line }x+2y=4\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&4&0\\ \hline y&0&2\\ \hline \end{array}
\displaystyle \text{So, the line passes through }(4,0)\text{ and }(0,2).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+2y\leq4,\text{ we get }0+0\leq4\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }3x+y=6\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&2&0\\ \hline y&0&6\\ \hline \end{array}
\displaystyle \text{So, the line passes through }(2,0)\text{ and }(0,6).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x+y\leq6,\text{ we get }0+0\leq6\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }x+y=4\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&4&0\\ \hline y&0&4\\ \hline \end{array}
\displaystyle \text{The line passes through }(4,0)\text{ and }(0,4).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }x+y\leq4,\text{ we get }0+0\leq4\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Also, }x\geq0,\ y\geq0,\text{ so the region lies in the Ist quadrant.}\displaystyle \text{Clearly, the feasible region is }OABCO.
\displaystyle \text{The intersection point of lines (i) and (ii) is }B\left(\frac{8}{5},\frac{6}{5}\right).
\displaystyle \text{Thus, the corner points are }O(0,0),\ A(2,0),\ B\left(\frac{8}{5},\frac{6}{5}\right)\text{ and }C(0,2).
\displaystyle \text{The values of }Z\text{ at corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=2x+5y\\ \hline O(0,0) & 0+0=0\\ \hline A(2,0) & 2\times2+5\times0=4\\ \hline B\left(\frac{8}{5},\frac{6}{5}\right) & 2\times\frac{8}{5}+5\times\frac{6}{5}=\frac{46}{5}=9.2\\ \hline C(0,2) & 2\times0+5\times2=10\ \text{(Maximum)}\\ \hline \end{array}
\displaystyle \text{Hence, the maximum value of }Z\text{ is }10\text{ at }C(0,2).
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\displaystyle \textbf{Question 24. }\text{Find the maximum value of the objective function } \\ Z=8x+9y   \\ \text{Subject to constraints } \\ 2x+3y\leq6,\ 3x-2y\leq6,\ y\leq1,\ x\geq0,   y\geq0. \hspace{0.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have the following LPP,}
\displaystyle \text{Maximize }Z=8x+9y
\displaystyle \text{Subject to the constraints}
\displaystyle 2x+3y\leq6,\quad 3x-2y\leq6,\quad y\leq1\text{ and }x,\ y\geq0
\displaystyle \text{Now, considering the inequations as equations, we get}
\displaystyle 2x+3y=6\qquad \cdots(i)
\displaystyle 3x-2y=6\qquad \cdots(ii)
\displaystyle y=1\qquad \cdots(iii)
\displaystyle \text{Table for line }2x+3y=6\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&3&0\\ \hline y&0&2\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(3,0)\text{ and }(0,2).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }2x+3y\leq6,\text{ we get }0\leq6\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{Table for line }3x-2y=6\text{ is}
\displaystyle \begin{array}{|c|cc|} \hline x&2&0\\ \hline y&0&-3\\ \hline \end{array}
\displaystyle \text{So, it passes through the points }(2,0)\text{ and }(0,-3).
\displaystyle \text{On putting }(0,0)\text{ in the inequality }3x-2y\leq6,\text{ we get }0\leq6\text{ (which is true).}
\displaystyle \text{So, the half plane is towards the origin.}
\displaystyle \text{The line }y=1\text{ is perpendicular to Y-axis.}
\displaystyle \text{On putting }(0,0)\text{ in the inequality }y\leq1,\text{ we get }0\leq1\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 quadrant.}
\displaystyle \text{The point of intersection of Eqs. (i) and (ii) is }\left(\frac{30}{13},\frac{6}{13}\right),
\displaystyle \text{Eqs. (ii) and (iii) is }\left(\frac{8}{3},1\right)\text{ and Eqs. (i) and (iii) is }\left(\frac{3}{2},1\right).\displaystyle \text{Clearly, feasible region is }OABCDO,\text{ whose corner points are }O(0,0),\ A(2,0),
\displaystyle B\left(\frac{30}{13},\frac{6}{13}\right),\ C\left(\frac{3}{2},1\right)\text{ and }D(0,1).
\displaystyle \text{The values of }Z\text{ at corner points are as follows}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner points} & \text{Value of }Z=8x+9y\\ \hline O(0,0) & 0+0=0\\ \hline A(2,0) & 8\times2+0=16\\ \hline B\left(\frac{30}{13},\frac{6}{13}\right) & 8\times\frac{30}{13}+9\times\frac{6}{13}=\frac{294}{13}=22.62\ \text{(Maximum)}\\ \hline C\left(\frac{3}{2},1\right) & 8\times\frac{3}{2}+9\times1=21\\ \hline D(0,1) & 0+9\times 1=9\\ \hline \end{array}
\displaystyle \text{In the table, we find that maximum value of }Z\text{ is }22.62,\text{ at }B\left(\frac{30}{13},\frac{6}{13}\right).
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