NCERT Solutions Class 12 Maths (Linear Programming) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-12 (Linear Programming) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1. Refer to Example 9. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Answer. $\begin{array}{c}\text{Let the diet contain}x\text{and}y\text{packets of foods}P\text{and}Q\text{respectively. Therefore}\\ x\ge 0\text{and}y\ge 0\\ \text{The mathematical formulation of the given problem is as follows.}\\ \text{Maximize}z=6x+3y\dots \left(1\right)\\ \text{subject to the constraints,}\end{array}$ $\begin{array}{c}4x+y\ge 80\\ x+5y\ge 115\\ 3x+2y\le 150\\ x,y\ge 0\\ \text{The feasible region determined by the system of constraints is as follows.}\end{array}$  The corner points of feasible region are A (15, 20), B (40, 15), and C (2, 72). The values of at these corner points are as follows.  Thus, the maximum value Otz is 28S at (40, 15). Therefore, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet is 285 unit.

Q2. A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 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 Rs 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?

Answer. Let the farmer mix x bags Of brand p and v bags Of brand Q. The given information can be compiled in a table as follows.  $\begin{array}{c}\text{The given problem can be formulated as follows.}\\ \text{Minimize}z=250x+200y\dots \left(1\right)\\ \text{subject to the constraints,}\\ 3x+1.5y\ge 18\\ 2.5x+11.25y\ge 45\\ 2x+3y\ge 24\\ x,y\ge 0\\ \text{The feasible region determined by the system of constraints is as follows.}\end{array}$  The corner points Of the feasible region are A (18, O), B (9, 2), C (3, 6), and D (O, 12). The values ot z at these corner points are as follows.  $\begin{array}{c}\text{As the feasible region is unbounded, therefore, 1950 may or may not be the minimum}\\ \text{value of}z\text{.}\\ \text{For this, we draw a graph of the inequality,}250x+200y<1950\text{or}5x+4y<39,\text{and}\\ \text{check whether the resulting half plane has points in common with the feasible region or}\\ \text{not.}\end{array}$ $\begin{array}{c}\text{It can be seen that the feasible region has no common point with}5x+4y<39\\ \text{Therefore, the minimum value of}z\text{is}2000\text{at}(3,6)\text{.}\\ \text{Thus,}3\text{bags of brand}P\text{and}6\text{bags of brand}Q\text{should be used in the mixture to}\\ \text{minimize the cost to Rs}1950\text{.}\end{array}$

Q3. 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:  One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Answer. $\begin{array}{c}\text{Let the mixture contain}x\text{kg of food}x\text{and}y\text{kg of food}Y\text{.}\\ \text{The mathematical formulation of the given problem is as follows.}\\ \text{Minimize}z=16x+20y\dots \left(1\right)\\ \text{subject to the constraints,}\end{array}$ $\begin{array}{c}x+2y\ge 10\\ x+y\ge 6\\ 3x+y\ge 8\\ x,y\ge 0\end{array}$  The corner points of the feasible region are A (10, O), B (2, 4), C (I, 5), and D (O, 8). The values of z at these corner points are as follows.  $\begin{array}{c}\text{As the feasible region is unbounded, therefore,}112\text{may or may not be the minimum}\\ \text{value of}z\\ \text{For this, we draw a graph of the inequality,}16x+20y<112\text{or}4x+5y<28,\text{and}\\ \text{check whether the resulting half plane has points in common with the feasible region or}\\ \text{not.}\end{array}$ $\begin{array}{c}\text{It can be seen that the feasible region has no common point with}4x+5y<28\\ \text{Therefore, the minimum value of}z\text{is}112\text{at}(2,4)\text{.}\\ \text{Thus, the mixture should contain}2kg\text{of food}x\text{and}4kg\text{of food}Y\text{. The minimum cost of}\\ \text{the mixture is Rs}112\text{.}\end{array}$

Q4. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:  Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Answer. $\begin{array}{c}\text{Let}x\text{and}y\text{toys of type}A\text{and type}B\text{respectively be manufactured in a day.}\\ \text{The given problem can be formulated as follows.}\\ \text{Maximize}z=7.5x+5y\dots \text{(1)}\\ \text{subject to the constraints,}\\ 2x+y\le 60\\ x\le 20\\ x\le 20\\ x,y\ge 0\end{array}$  The corner points of the feasible region are A (20. 0). B (20. 20). C (15, 30). and 0 (0, 40). The values of z at these corner points are as follows.  The maximum value of z s 262.5 at (15 , 30). Thus, the manufacturer should manufacture 15 toys of type A and 30 toys Of type to maximize the profit.

Q5. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

Answer. $\begin{array}{c}\text{Let the airline sell}x\text{tickets of executive class and}y\text{tickets of economy class.}\\ \text{The mathematical formulation of the given problem is as follows.}\\ \text{Maximize}z=1000x+600y\dots \left(1\right)\\ \text{subject to the constraints,}\\ x+y\le 200\\ x\ge 20\\ x\ge 20\\ y-4x\ge 0\\ x,y\ge 0\end{array}$  The corner points Of the feasible region are A (20, 80), B (40, 160), and c (20, 180). The values of z at these corner points are as follows.  The maximum value Of z is 136000 at (40, 160). Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.

Q6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:  How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Answer. $\begin{array}{c}\text{Let godown A supply}x\text{and}y\text{quintals of grain to the shops}D\text{and}E\text{respectively. Then,}\\ (100-x-y)\text{will be supplied to shop F.}\\ \text{The requirement at shop D is 60 quintals since}x\text{quintals are transported from godown}\end{array}$ $\begin{array}{c}\text{A. Therefore, the remaining}(60-x)\text{quintals will be transported from godown B.}\\ \text{Similarly,}(50-y)\text{quintals and}40-(100-x-y)=(x+y-60)\text{quintals will be}\\ \text{transported from godown}8\text{to shop}E\text{and}F\text{respectively.}\\ \text{The given problem can be represented diagrammatically as follows.}\end{array}$  $\begin{array}{c}60-x\ge 0,50-y\ge 0,\text{and}x+y-60\ge 0\\ \Rightarrow x\le 60,y\le 50,\text{and}x+y\ge 60\\ \text{Total transportation cost}z\text{is given by,}\\ z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)\\ =6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180\\ =2.5x+1.5y+410\end{array}$ $\begin{array}{c}\text{The given problem can be formulated as}\\ \text{Minimize}z=2.5x+1.5y+410\dots \text{(1)}\\ \text{subject to the constraints,}\\ x+y\le 100\\ x+y\le 100\\ x\le 60\\ y\le 50\\ y\le 50\\ x+y\ge 0\\ \text{The feasible region determined by the system of constraints is as follows.}\end{array}$  The corner points are A (60, 0), B (60, 40), C (SO, 50), and D (10, SO). The values of z at these corner points are as follows.  The minimum value of is 510 at (10, 50). Thus, the amount of grain transported from A to D, E, and F is 10 quintals, SO quintals, and 40 quintals respectively and from B to D. E, and F is 50 0 quintals, and 0

Q7. An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:  Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Answer. Let x and V litres Of Oil be supplied from A to the petrol pumps, D and E. Then, (7000 — x — y) will supplied from A to petrol pump F. The requirement at petrol pump D is 4500 L. Since x L are transported from depot A. the remaining (4500 —x) will be transported from petrol pump B. Similarly. (3000 - V) L and 3500 - (7000 - x - y) - (x y - 3500) L will be transported from depot B to petrol pump E and F respectively. The given problem can be represented diagrammatically as follows.  $\begin{array}{c}x\ge 0,y\ge 0,\text{and}(7000-x-y)\ge 0\\ \Rightarrow x\ge 0,y\ge 0,\text{and}x+y\le 7000\\ 4500-x\ge 0,3000-y\ge 0,\text{and}x+y-3500\ge 0\\ \Rightarrow x\le 4500,y\le 3000,\text{and}x+y\ge 3500\\ \text{cost of transporting}10\text{L of petrol}=\text{Re}1\end{array}$ $\begin{array}{c}\text{Therefore, total transportation cost is given by,}\\ z=\frac{7}{10}\times x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)\\ =0.3x+0.1y+3950\end{array}$ $\begin{array}{c}\text{The problem can be formulated as follows.}\\ \text{Minimize}z=0.3x+0.1y+3950\dots \left(1\right)\\ \text{subject to the constraints,}\\ x+y\le 7000\\ x\le 4500\\ x\le 4500\\ y\le 3000\\ x+y\ge 0\\ x,y\ge 0\end{array}$  $\begin{array}{c}\text{The corner points of the feasible region are}A(3500,0),\text{B}(4500,0),C(4500,2500),\text{D}\\ (4000,3000),\text{and}E(500,3000).\\ \text{The values of}z\text{at these corner points are as follows.}\end{array}$  The minimum value of is 4400 at (500, 3000). Thus, the oil supplied from depot A is 500 L 3000 and 3500 L and from depot B is 4000 L, O L, and O L to petrol pumps D. E, and F respectively. The minimum transportation cost iS RS 4400.

Q8. A fruit grower can use two types of fertilizer in his garden, brand P and brand 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.  If the grower wants to minimise 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?

Answer. $\begin{array}{c}\text{Let the fruit grower use}x\text{bags of brand}P\text{and}y\text{bags of brand Q.}\\ \text{The problem can be formulated as follows.}\\ \text{Minimize}z=3x+3.5y\dots \left(1\right)\\ \text{subiect to the constraints.}\\ x+2y\ge 240\\ x+2.5y\ge 90\\ 1.5x+2y\le 310\\ x,y\ge 0\end{array}$  $\begin{array}{c}\text{The maximum value of}z\text{is}470\text{at}(40,100).\\ \text{Thus,}40\text{bags of brand}P\text{and}100\text{bags of brand}Q\text{should be added to the garden to}\\ \text{minimize the amount of nitrogen.}\\ \text{The minimum amount of nitrogen added to the garden is}470kg.\end{array}$

Q9. Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

Answer. $\begin{array}{c}\text{Let the fruit grower use}x\text{bags of brand}P\text{and}y\text{bags of brand Q.}\\ \text{The problem can be formulated as follows.}\\ \text{Maximize}z=3x+3.5y\dots \left(1\right)\\ \text{subject to the constraints,}\end{array}$ $\begin{array}{cc}x+2y\ge 240& \dots \left(2\right)\\ x+0.5y\ge 90& \dots \left(3\right)\\ 1.5x+2y\le 310& \dots \left(4\right)\\ x,y\ge 0& \dots \left(5\right)\\ \text{The feasible region determined by the system of constraints is as follows.}& \end{array}$  $\begin{array}{c}\text{The maximum value of}z\text{is}595\text{at}(140,50)\text{.}\\ \text{Thus,}140\text{bags of brand}P\text{and}50\text{bags of brand}Q\text{should be used to maximize the}\\ \text{amount of nitrogen.}\\ \text{The maximum amount of nitrogen added to the garden is}595kg.\end{array}$

Q10. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

Answer. $\begin{array}{c}\text{Let}x\text{and}y\text{be the number of dolls of type}A\text{and}B\text{respectively that are produced per}\\ \text{week.}\\ \text{The given problem can be formulated as follows.}\\ \text{Maximize}z=12x+16y\dots \left(1\right)\\ \text{subject to the constraints,}\end{array}$ $\begin{array}{cc}x+y\le 1200& \dots \left(2\right)\\ y\le \frac{x}{2}\Rightarrow x\ge 2y& \dots \left(3\right)\\ x-3y\le 600& \dots \left(4\right)\\ x,y\ge 0& \dots \left(5\right)\end{array}$  $\begin{array}{c}\text{The maximum value of}z\text{is}16000\text{at}(800,400).\\ \text{Thus,}800\text{and}400\text{dolls of type}A\text{and type}B\text{should be produced respectively to get the}\\ \text{maximum profit of Rs}16000.\end{array}$