\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.}

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