Question 1

A diet of two foods F₁ and F₂ contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:

\displaystyle \begin{array}{|c|c|c|} \hline \text{Nutrients} & F_1 & F_2\\ \hline \text{Thiamine} & 0.25 & 0.10\\ \hline \text{Phosphorous} & 0.75 & 1.50\\ \hline \text{Iron} & 1.60 & 0.80\\ \hline \end{array}

The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron. The cost of F₁ is 20 paise per 25 gms while the cost of F₂ is 15 paise per 25 gms. Find the minimum cost of diet.

\displaystyle \text{Answer:}
\displaystyle \text{The information above can be expressed in the following table:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Food}/\text{Nutrients} & \text{Thiamine} & \text{Phosphorous} & \text{Iron} & \text{Price per 25 g (Rs)}\\ \hline F_1 & 0.25 & 0.75 & 1.60 & 0.20\\ \hline F_2 & 0.10 & 1.50 & 0.80 & 0.15\\ \hline \text{Minimum Requirement} & 1.00 & 7.50 & 10.00 & {}\\ \hline \end{array}
\displaystyle \text{Let the amount of food }F_1\text{ and }F_2\text{ required be }x\text{ and }y\text{ units respectively.}
\displaystyle \text{Cost of }F_1 = 0.20x
\displaystyle \text{Cost of }F_2 = 0.15y
\displaystyle \text{So, cost of diet }Z = 0.20x + 0.15y
\displaystyle \text{Now,}
\displaystyle 0.25x + 0.10y \ge 1.00
\displaystyle \text{i.e. the minimum requirement of Thiamine should be }1.00\text{ mg from both foods }F_1\text{ and }F_2.
\displaystyle 0.75x + 1.50y \ge 7.50
\displaystyle \text{i.e. the minimum requirement of Phosphorous should be }7.50\text{ mg from both foods }F_1\text{ and }F_2.
\displaystyle 1.60x + 0.80y \ge 10.00
\displaystyle \text{i.e. the minimum requirement of Iron should be }10.00\text{ mg from both foods }F_1\text{ and }F_2.
\displaystyle \text{Hence, mathematical formulation of LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ which minimise }Z = 0.20x + 0.15y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }0.25x + 0.10y \ge 1.00
\displaystyle \text{(ii) }0.75x + 1.50y \ge 7.50
\displaystyle \text{(iii) }1.60x + 0.80y \ge 10.00
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0\;(\text{quantity cannot be negative})
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z\\ \hline A(10,0) & 2\\ \hline B(5,2.5) & 1.375\\ \hline C(0,12.5) & 1.875\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }B(5,2.5).
\displaystyle \text{Hence, the minimum cost of diet is Rs }1.375.

Question 2

A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B are available at a cost of ₹4 and ₹3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?             [CBSE 2004]

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Food}/\text{Nutrients} & \text{Vitamins} & \text{Minerals} & \text{Calories} & \text{Price per unit (Rs)}\\ \hline \text{Food A} & 200 & 1 & 40 & 4\\ \hline \text{Food B} & 100 & 2 & 40 & 3\\ \hline \text{Minimum Requirement} & 4000 & 50 & 1400 & {}\\ \hline \end{array}
\displaystyle \text{Let the quantities of the foods be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Cost of Food A }= 4x
\displaystyle \text{Cost of Food B }= 3y
\displaystyle \text{Total cost of the combination }Z = 4x + 3y
\displaystyle \text{Now,}
\displaystyle 200x + 100y \ge 4000
\displaystyle \text{i.e. the minimum requirement of vitamins from the two foods should be }4000.
\displaystyle x + 2y \ge 50
\displaystyle \text{i.e. the minimum requirement of minerals from the two foods should be }50.
\displaystyle 40x + 40y \ge 1400
\displaystyle \text{i.e. the minimum requirement of calories from the two foods should be }1400.
\displaystyle \text{Hence, mathematical formulation of LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ which minimise }Z = 4x + 3y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }200x + 100y \ge 4000
\displaystyle \text{(ii) }x + 2y \ge 50
\displaystyle \text{(iii) }40x + 40y \ge 1400
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 4x + 3y\\ \hline A(0,40) & 120\\ \hline B(5,30) & 110\\ \hline C(20,15) & 125\\ \hline D(50,0) & 200\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }B(5,30).
\displaystyle \text{Hence, the minimum cost of foods is Rs }110.

Question 3

To maintain one’s health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Nutrients} & \text{Food I} & \text{Food II} & \text{Minimum daily requirement}\\ \hline \text{Calcium} & 10 & 4 & 20\\ \hline \text{Protein} & 5 & 6 & 20\\ \hline \text{Calories} & 2 & 6 & 12\\ \hline \text{Price} & \text{Rs }0.60\ \text{per unit} & \text{Rs }1.00\ \text{per unit} & {}\\ \hline \end{array}

Find the combination of food items so that the cost may be minimum.

\displaystyle \text{Answer:}
\displaystyle \text{Let the quantities of foods chosen be }x\text{ and }y.
\displaystyle \text{Cost of food }X = 0.6x
\displaystyle \text{Cost of food }Y = y
\displaystyle \text{Cost of diet }Z = 0.6x + y
\displaystyle \text{Now,}
\displaystyle 10x + 4y \ge 20
\displaystyle \text{i.e. the minimum daily requirement of calcium in the diet is }20\text{ units.}
\displaystyle 5x + 6y \ge 20
\displaystyle \text{i.e. the minimum daily requirement of protein in the diet is }20\text{ units.}
\displaystyle 2x + 6y \ge 12
\displaystyle \text{i.e. the minimum daily requirement of calories in the diet is }12\text{ units.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ such that }Z\text{ is minimised.}
\displaystyle Z = 0.6x + y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }10x + 4y \ge 20
\displaystyle \text{(ii) }5x + 6y \ge 20
\displaystyle \text{(iii) }2x + 6y \ge 12
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 0.6x + y\\ \hline A(0,5) & 5\\ \hline B(1,2.5) & 3.1\\ \hline C\!\left(\frac{8}{3},\frac{10}{9}\right) & 2.712\\ \hline D(6,0) & 3.6\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }C\!\left(\frac{8}{3},\frac{10}{9}\right).
\displaystyle \text{Hence, the minimum value of }Z\text{ is Rs }2.712.

Question 4

A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of A contains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed using the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & \text{Food A} & \text{Food B} & \text{Minimum daily requirement}\\ \hline \text{Thiamine} & 0.12 & 0.10 & 0.5\\ \hline \text{Calories} & 100 & 150 & 600\\ \hline \text{Price} & 0.10\text{ per unit} & 0.10\text{ per unit} & {}\\ \hline \end{array}
\displaystyle \text{Let the quantities of foods A and B be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Cost of food A }= 0.10x
\displaystyle \text{Cost of food B }= 0.10y
\displaystyle \text{Cost of diet }Z = 0.10x + 0.10y
\displaystyle \text{Now,}
\displaystyle 0.12x + 0.10y \ge 0.5
\displaystyle \text{i.e. the minimum requirement of thiamine in the foods is }0.5\text{ mg.}
\displaystyle 100x + 150y \ge 600
\displaystyle \text{i.e. the minimum requirement of calories in the foods is }600.
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 0.10x + 0.10y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }0.12x + 0.10y \ge 0.5
\displaystyle \text{(ii) }100x + 150y \ge 600
\displaystyle \text{i.e. }2x + 3y \ge 12
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 0.10x + 0.10y\\ \hline A\!\left(1.875,2.75\right) & 0.4625\\ \hline B(6,0) & 0.6\\ \hline C(0,5) & 0.5\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }A\!\left(1.875,2.75\right).
\displaystyle \text{Hence, the minimum cost of the foods is Rs }0.4625.

Question 5

A dietician mixed together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:

\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Food} & \text{Vitamin A} & \text{Vitamin B} & \text{Vitamin C} & \text{Vitamin D}\\ \hline \text{Food X} & 1 & 1 & 1 & 2\\ \hline \text{Food Y} & 2 & 1 & 3 & 1\\ \hline \end{array}

One kg of food X costs ₹5, whereas one kg of food Y costs ₹8. Find the least cost of the mixture which will produce the desired diet.

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed with the help of the following table:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Food}/\text{Nutrients} & \text{Vitamin A} & \text{Vitamin B} & \text{Vitamin C} & \text{Vitamin D} & \text{Price per kg (Rs)}\\ \hline \text{Food X} & 1 & 1 & 1 & 2 & 5\\ \hline \text{Food Y} & 2 & 1 & 3 & 1 & 8\\ \hline \text{Minimum Requirement} & 6 & 7 & 11 & 9 & {}\\ \hline \end{array}
\displaystyle \text{Let the quantities of foods X and Y be }x\text{ and }y.
\displaystyle \text{Cost of food X }= 5x
\displaystyle \text{Cost of food Y }= 8y
\displaystyle \text{Cost of the meal }Z = 5x + 8y
\displaystyle \text{Now,}
\displaystyle x + 2y \ge 6
\displaystyle \text{i.e. the minimum requirement of Vitamin A in the foods X and Y is }6\text{ units.}
\displaystyle x + y \ge 7
\displaystyle \text{i.e. the minimum requirement of Vitamin B in the two foods is }7\text{ units.}
\displaystyle x + 3y \ge 11
\displaystyle \text{i.e. the minimum requirement of Vitamin C in the two foods is }11\text{ units.}
\displaystyle 2x + y \ge 9
\displaystyle \text{i.e. the minimum requirement of Vitamin D in the foods is }9\text{ units.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 5x + 8y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }x + 2y \ge 6
\displaystyle \text{(ii) }x + y \ge 7
\displaystyle \text{(iii) }x + 3y \ge 11
\displaystyle \text{(iv) }2x + y \ge 9
\displaystyle \text{(v) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 5x + 8y\\ \hline A(0,9) & 72\\ \hline B(2,5) & 50\\ \hline C(5,2) & 41\\ \hline D(11,0) & 55\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }C(5,2).
\displaystyle \text{Hence, the minimum cost of the diet is Rs }41.

Question 6

A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs ₹4 per unit and F₂ costs ₹6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.                    [CBSE 2009]

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & F_1 & F_2 & \text{Minimum daily requirement}\\ \hline \text{Vitamins} & 3 & 6 & 80\\ \hline \text{Minerals} & 4 & 3 & 100\\ \hline \text{Price} & 4\text{ per unit} & 6\text{ per unit} & {}\\ \hline \end{array}
\displaystyle \text{Let the quantities of foods }F_1\text{ and }F_2\text{ be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Cost of food }F_1 = 4x
\displaystyle \text{Cost of food }F_2 = 6y
\displaystyle \text{Cost of diet }Z = 4x + 6y
\displaystyle \text{Now,}
\displaystyle 3x + 6y \ge 80
\displaystyle \text{i.e. the minimum requirement of vitamins from the two foods is }80\text{ units.}
\displaystyle 4x + 3y \ge 100
\displaystyle \text{i.e. the minimum requirement of minerals from the two foods is }100\text{ units.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 4x + 6y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }3x + 6y \ge 80
\displaystyle \text{(ii) }4x + 3y \ge 100
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 4x + 6y\\ \hline A\!\left(24,\frac{4}{3}\right) & 104\\ \hline B\!\left(0,\frac{100}{3}\right) & 200\\ \hline C\!\left(\frac{80}{3},0\right) & 106.667\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }A\!\left(24,\frac{4}{3}\right).
\displaystyle \text{Hence, the minimum cost of diet is Rs }104.

Question 7

Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs ₹5 per kg and rice costs ₹4 per kg.          [CBSE 2002]

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed using the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & \text{Bran} & \text{Rice} & \text{Minimum requirement}\\ \hline \text{Proteins (g)} & 80 & 100 & 88\\ \hline \text{Iron (mg)} & 40 & 30 & 36\\ \hline \text{Price} & 5\text{ per kg} & 4\text{ per kg} & {}\\ \hline \end{array}
\displaystyle \text{Let the amounts of Bran and Rice required be }x\text{ kg and }y\text{ kg respectively.}
\displaystyle \text{Cost of Bran }= 5x
\displaystyle \text{Cost of Rice }= 4y
\displaystyle \text{Cost of the cereal }Z = 5x + 4y
\displaystyle \text{Now,}
\displaystyle 80x + 100y \ge 88
\displaystyle \text{i.e. the minimum requirement of protein in the cereal is }88\text{ g.}
\displaystyle 40x + 30y \ge 36
\displaystyle \text{i.e. the minimum requirement of iron in the cereal is }36\text{ mg.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 5x + 4y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }80x + 100y \ge 88
\displaystyle \text{(ii) }40x + 30y \ge 36
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 5x + 4y\\ \hline A(0,1.2) & 4.8\\ \hline B(0.6,0.4) & 4.6\\ \hline C(1.1,0) & 5.5\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }B(0.6,0.4).
\displaystyle \text{Hence, the minimum cost of the cereal is Rs }4.6.

Question 8

A wholesale dealer deals in two kinds, A and B (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture A costs ₹8 per kg and mixture B costs ₹12 per kg. Assuming that mixtures A and B are uniform, use graphical method to determine the number of kg of each mixture which he should use to minimise the cost of the bag.

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & \text{Bag A} & \text{Bag B} & \text{Minimum Requirement (g)}\\ \hline \text{Almonds (g)} & 60 & 30 & 240\\ \hline \text{Cashew Nuts (g)} & 30 & 60 & 300\\ \hline \text{Hazel Nuts (g)} & 30 & 180 & 540\\ \hline \text{Price} & 8\text{ per kg} & 12\text{ per kg} & {}\\ \hline \end{array}
\displaystyle \text{Let the numbers of bags chosen of A and B be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Cost of Bag A }= 8x
\displaystyle \text{Cost of Bag B }= 12y
\displaystyle \text{Total cost of bags }Z = 8x + 12y
\displaystyle \text{Now,}
\displaystyle 60x + 30y \ge 240
\displaystyle \text{i.e. the minimum requirement of almonds from both the bags is }240\text{ g.}
\displaystyle 30x + 60y \ge 300
\displaystyle \text{i.e. the minimum requirement of cashew nuts from both the bags is }300\text{ g.}
\displaystyle 30x + 180y \ge 540
\displaystyle \text{i.e. the minimum requirement of hazel nuts from both the bags is }540\text{ g.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 8x + 12y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }60x + 30y \ge 240
\displaystyle \text{i.e. }2x + y \ge 8
\displaystyle \text{(ii) }30x + 60y \ge 300
\displaystyle \text{i.e. }x + 2y \ge 10
\displaystyle \text{(iii) }30x + 180y \ge 540
\displaystyle \text{i.e. }x + 6y \ge 18
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 8x + 12y\\ \hline A(18,0) & 144\\ \hline B(6,2) & 72\\ \hline C(2,4) & 64\\ \hline D(0,8) & 96\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }C(2,4).
\displaystyle \text{Hence, the minimum cost of the bags is Rs }64.

Question 9

One kind of cake requires 300 gm of flour and 15 gm of fat, another kind of cake requires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gm of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it as an LPP and solve it graphically.          [CBSE 2010]

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & \text{Cake 1} & \text{Cake 2} & \text{Maximum Availability (g)}\\ \hline \text{Flour (g)} & 300 & 150 & 7500\\ \hline \text{Fat (g)} & 15 & 30 & 600\\ \hline \text{Price} & 5\text{ per kg} & 4\text{ per kg} & {}\\ \hline \end{array}
\displaystyle \text{Let }x\text{ and }y\text{ units of cake 1 and cake 2 be made.}
\displaystyle \text{Number of cakes made }Z = x + y
\displaystyle \text{Now,}
\displaystyle 300x + 150y \le 7500
\displaystyle \text{i.e. the maximum availability of flour is }7500\text{ g for both cakes.}
\displaystyle 15x + 30y \le 600
\displaystyle \text{i.e. the maximum availability of fat is }600\text{ g for both cakes.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that maximise }Z = x + y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }300x + 150y \le 7500
\displaystyle \text{i.e. }2x + y \le 50
\displaystyle \text{(ii) }15x + 30y \le 600
\displaystyle \text{i.e. }x + 2y \le 40
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is bounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = x + y\\ \hline A(25,0) & 25\\ \hline B(20,10) & 30\\ \hline C(0,20) & 20\\ \hline O(0,0) & 0\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is maximised at }B(20,10).
\displaystyle \text{Hence, the maximum number of cakes that can be made is }30.

Question 10

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹60 kg and food Q costs ₹80 kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & P & Q & \text{Minimum requirement}\\ \hline \text{Vitamin A} & 3 & 4 & 8\\ \hline \text{Vitamin B} & 5 & 2 & 11\\ \hline \text{Price} & 60\text{ per kg} & 80\text{ per kg} & {}\\ \hline \end{array}
\displaystyle \text{Let the mixture contain }x\text{ kg and }y\text{ kg of foods P and Q respectively.}
\displaystyle \text{Cost of food P }= 60x
\displaystyle \text{Cost of food Q }= 80y
\displaystyle \text{Cost of mixture }Z = 60x + 80y
\displaystyle \text{Now,}
\displaystyle 3x + 4y \ge 8
\displaystyle \text{i.e. the minimum requirement of Vitamin A from the mixture of P and Q is }8\text{ units.}
\displaystyle 5x + 2y \ge 11
\displaystyle \text{i.e. the minimum requirement of Vitamin B from the mixture of P and Q is }11\text{ units.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 60x + 80y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }3x + 4y \ge 8
\displaystyle \text{(ii) }5x + 2y \ge 11
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 60x + 80y\\ \hline A(0,5.5) & 440\\ \hline B(2,0.5) & 160\\ \hline C\!\left(\frac{8}{3},0\right) & 160\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is minimised on the line joining points }B(2,0.5)\text{ and }C\!\left(\frac{8}{3},0\right).
\displaystyle \text{Hence, the minimum cost of mixture is Rs }160.

Question 11

One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & \text{Cake 1} & \text{Cake 2} & \text{Maximum Availability}\\ \hline \text{Flour (g)} & 200 & 100 & 5000\\ \hline \text{Fat (g)} & 25 & 50 & 1000\\ \hline \end{array}
\displaystyle \text{Let the numbers of Cake 1 and Cake 2 to be made be }x\text{ and }y.
\displaystyle \text{Number of cakes made }Z = x + y
\displaystyle \text{Now,}
\displaystyle 200x + 100y \le 5000
\displaystyle \text{i.e. the maximum flour available for both cakes combined is }5000\text{ g.}
\displaystyle 25x + 50y \le 1000
\displaystyle \text{i.e. the maximum fat available for both cakes combined is }1000\text{ g.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that maximise }Z = x + y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }200x + 100y \le 5000
\displaystyle \text{i.e. }2x + y \le 50
\displaystyle \text{(ii) }25x + 50y \le 1000
\displaystyle \text{i.e. }x + 2y \le 40
\displaystyle \text{(iii) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is bounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = x + y\\ \hline A(20,10) & 30\\ \hline B(0,20) & 20\\ \hline C(25,0) & 25\\ \hline O(0,0) & 0\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is maximised at }A(20,10).
\displaystyle \text{Hence, the maximum number of cakes that can be made is }30.

Question 12

A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed with the help of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & P & Q & \text{Requirement}\\ \hline \text{Calcium} & 12 & 3 & \text{At least }240\\ \hline \text{Iron} & 4 & 20 & \text{At least }460\\ \hline \text{Cholesterol} & 6 & 4 & \text{At most }300\\ \hline \text{Vitamin A} & 6 & 3 & {}\\ \hline \end{array}
\displaystyle \text{Let the numbers of packets bought of P and Q be }x\text{ and }y.
\displaystyle \text{Vitamin A from P }= 6x
\displaystyle \text{Vitamin A from Q }= 3y
\displaystyle \text{Vitamin A in the diet }Z = 6x + 3y
\displaystyle \text{Now,}
\displaystyle 12x + 3y \ge 240
\displaystyle \text{i.e. the minimum requirement of calcium in the diet from P and Q combined is }240\text{ units.}
\displaystyle 4x + 20y \ge 460
\displaystyle \text{i.e. the minimum requirement of iron from P and Q combined is }460\text{ units.}
\displaystyle 6x + 4y \le 300
\displaystyle \text{i.e. the maximum requirement of cholesterol from P and Q combined is }300\text{ units.}
\displaystyle \text{Hence, the mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 6x + 3y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }12x + 3y \ge 240
\displaystyle \text{i.e. }4x + y \ge 80
\displaystyle \text{(ii) }4x + 20y \ge 460
\displaystyle \text{i.e. }x + 5y \ge 115
\displaystyle \text{(iii) }6x + 4y \le 300
\displaystyle \text{i.e. }3x + 2y \le 150
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is bounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 6x + 3y\\ \hline A(15,20) & 150\\ \hline B(40,15) & 285\\ \hline C(2,72) & 228\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is minimised at }A(15,20).
\displaystyle \text{Hence, }15\text{ packets of P and }20\text{ packets of Q should be used.}
\displaystyle \text{The minimum amount of Vitamin A is }150\text{ units.}

Question 13

A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed with the help of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & P & Q & \text{Minimum Requirement}\\ \hline \text{Element A} & 3 & 1.5 & 18\\ \hline \text{Element B} & 2.5 & 11.25 & 45\\ \hline \text{Element C} & 2 & 3 & 24\\ \hline \text{Cost per bag} & 250 & 200 & {}\\ \hline \end{array}
\displaystyle \text{Let }x\text{ bags of P and }y\text{ bags of Q be bought.}
\displaystyle \text{Cost of P }= 250x
\displaystyle \text{Cost of Q }= 200y
\displaystyle \text{Cost of mixture }Z = 250x + 200y
\displaystyle \text{Now,}
\displaystyle 3x + 1.5y \ge 18
\displaystyle \text{i.e. the minimum requirement of element A from both P and Q combined is }18\text{ units.}
\displaystyle 2.5x + 11.25y \ge 45
\displaystyle \text{i.e. the minimum requirement of element B from both P and Q combined is }45\text{ units.}
\displaystyle 2x + 3y \ge 24
\displaystyle \text{i.e. the minimum requirement of element C from both P and Q combined is }24\text{ units.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 250x + 200y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }3x + 1.5y \ge 18
\displaystyle \text{(ii) }2.5x + 11.25y \ge 45
\displaystyle \text{(iii) }2x + 3y \ge 24
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 250x + 200y\\ \hline A(0,12) & 2400\\ \hline B(3,6) & 1950\\ \hline C(9,2) & 2650\\ \hline D(18,0) & 4500\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is minimised at }B(3,6).
\displaystyle \text{Hence, }3\text{ bags of P and }6\text{ bags of Q should be purchased.}
\displaystyle \text{The minimum cost of the mixture is Rs }1950.

Question 14

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. 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 & 2 & 3\\ \hline Y & 2 & 2 & 1\\ \hline \end{array}

One kg of food X costs ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet.

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed in the form of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & X & Y & \text{Minimum Requirement}\\ \hline \text{Vitamin A} & 1 & 2 & 10\\ \hline \text{Vitamin B} & 2 & 2 & 12\\ \hline \text{Vitamin C} & 3 & 1 & 8\\ \hline \text{Cost per kg} & 16 & 20 & {}\\ \hline \end{array}
\displaystyle \text{Let the quantities of X and Y purchased be }x\text{ kg and }y\text{ kg respectively.}
\displaystyle \text{Cost of X }= 16x
\displaystyle \text{Cost of Y }= 20y
\displaystyle \text{Cost of the mixture }Z = 16x + 20y
\displaystyle \text{Now,}
\displaystyle x + 2y \ge 10
\displaystyle \text{i.e. the minimum requirement of Vitamin A from the mixture of X and Y is }10\text{ units.}
\displaystyle 2x + 2y \ge 12
\displaystyle \text{i.e. the minimum requirement of Vitamin B from the mixture of X and Y is }12\text{ units.}
\displaystyle 3x + y \ge 8
\displaystyle \text{i.e. the minimum requirement of Vitamin C from the mixture of X and Y is }8\text{ units.}
\displaystyle \text{Hence, the mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 16x + 20y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }x + 2y \ge 10
\displaystyle \text{(ii) }2x + 2y \ge 12
\displaystyle \text{i.e. }x + y \ge 6
\displaystyle \text{(iii) }3x + y \ge 8
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is unbounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 16x + 20y\\ \hline A(0,8) & 160\\ \hline B(1,5) & 116\\ \hline C(2,4) & 112\\ \hline D(6,2) & 136\\ \hline E(10,0) & 160\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is smallest at }C(2,4).
\displaystyle \text{Hence, the minimum cost of the mixture is Rs }112.

Question 15

A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

\displaystyle \begin{array}{|c|c|c|} \hline \text{kg per bag} & \text{Brand P} & \text{Brand Q}\\ \hline \text{Nitrogen} & 3 & 3.5\\ \hline \text{Phosphoric acid} & 1 & 2\\ \hline \text{Potash} & 3 & 1.5\\ \hline \text{Chlorine} & 1.5 & 2\\ \hline \end{array}

If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

\displaystyle \text{Answer:}
\displaystyle \text{The above information can be expressed with the help of the following table:}
\displaystyle \begin{array}{|c|c|c|c|} \hline {} & P & Q & \text{Requirement}\\ \hline \text{Phosphoric Acid} & 1 & 2 & \text{At least }240\\ \hline \text{Potash} & 3 & 1.5 & \text{At least }270\\ \hline \text{Chlorine} & 1.5 & 2 & \text{At most }310\\ \hline \text{Nitrogen} & 3 & 3.5 & {}\\ \hline \end{array}
\displaystyle \text{Let the numbers of bags of P and Q chosen be }x\text{ and }y.
\displaystyle \text{Nitrogen from P }= 3x
\displaystyle \text{Nitrogen from Q }= 3.5y
\displaystyle \text{Nitrogen from the mixture }Z = 3x + 3.5y
\displaystyle \text{Now,}
\displaystyle x + 2y \ge 240
\displaystyle \text{i.e. the minimum requirement of phosphoric acid in the mixture of P and Q is }240\text{ kg.}
\displaystyle 3x + 1.5y \ge 270
\displaystyle \text{i.e. the minimum requirement of potash in the mixture of P and Q is }270\text{ kg.}
\displaystyle 1.5x + 2y \le 310
\displaystyle \text{i.e. the maximum requirement of chlorine in the mixture of P and Q is }310\text{ kg.}
\displaystyle \text{Hence, mathematical formulation of the LPP is as follows:}
\displaystyle \text{Find }x\text{ and }y\text{ that minimise }Z = 3x + 3.5y
\displaystyle \text{Subject to the following constraints:}
\displaystyle \text{(i) }x + 2y \ge 240
\displaystyle \text{(ii) }3x + 1.5y \ge 270
\displaystyle \text{(iii) }1.5x + 2y \le 310
\displaystyle \text{(iv) }x \ge 0,\; y \ge 0
\displaystyle \text{The feasible region is bounded.}
\displaystyle \text{The corner points of the feasible region are as follows:}
\displaystyle \begin{array}{|c|c|} \hline \text{Point} & \text{Value of }Z = 3x + 3.5y\\ \hline A(20,140) & 550\\ \hline B(40,100) & 470\\ \hline C(140,50) & 595\\ \hline \end{array}
\displaystyle \text{So, }Z\text{ is minimised at }B(40,100).
\displaystyle \text{Hence, the minimum amount of nitrogen in the mixture is }470\text{ kg.}

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