# NCERT Solutions for Miscellaneous Exercise Chapter 12 Class 12 - Linear Programming

NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives more knowledge of solving a few types of linear programming problems using graphical methods. Miscellaneous exercise chapter 12 Class 12 are solved in a detailed manner with necessary steps and graphs. Going through NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives a better understanding of the linear programming problems. The questions discussed in the NCERT book Class 12 Maths chapter 12 miscellaneous solutions are a bit higher level as compared to the other two exercises. Class 12 Maths chapter 12 miscellaneous solutions are important for Class 12 CBSE Board Exams.

• Linear Programming Exercise 12.1

• Linear Programming Exercise 12.1

## Linear Programming Class 12 Chapter12-Miscellaneous Exercise

Question:1 Reference of Example 9 (Diet problem): 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 maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Let diet contain x packets of food P and y packets of food Q. Thus, $x\geq 0,y\geq 0$ .

The mathematical formulation of the given problem is as follows:

Total cost is Z . $Z=6x+3y$

Subject to constraint,

$4x+y\geq 80$

$x+5y\geq 115$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(15,20),B(40,15),C(2,72)$

The value of Z at corner points is as shown :

 corner points $Z=6x+3y$ $A(15,20)$ 150 MINIMUM $B(40,15)$ 285 maximum $C(2,72)$ 228

Hence, Z has a maximum value of 285 at the point $B(40,15)$ .

to maximise 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 is 285 units.

Question:2 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?

Let farmer mix x bags of brand P and y bags of brand Q. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 Vitamin A Vitamin B Cost Food P 3 5 60 Food Q 4 2 80 requirement 8 11

The given problem can be formulated as follows:

Therefore, we have

$3x+1.5y\geq 18$

$2.5x+11.25y\geq 45$

$2x+3y\geq 24$

$Z=250x+200y$

Subject to constraint,

$3x+1.5y\geq 18$

$2.5x+11.25y\geq 45$

$2x+3y\geq 24$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(18,0),B(9,2),C(3,6),D(0,12)$

The value of Z at corner points is as shown :

 corner points $Z=250x+200y$ $A(18,0)$ 4500 $B(9,2)$ 2650 $C(3,6)$ 1950 minimum $D(0,12)$ 2400

Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw $250x+200y< 1950$ and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with $250x+200y< 1950$ .

Hence, Z has a minimum value 1950 at point $C(3,6)$ .

Question:3 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:

 Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1

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?

Let mixture contain x kg of food X and y kg of food Y.

Mathematical formulation of given problem is as follows:

Minimize : $z=16x+20y$

Subject to constraint ,

$x+2y\geq 10$

$x+y\geq 6$

$3x+y\geq 8$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(10,0),B(2,4),C(1,5),D(0,8)$

The value of Z at corner points is as shown :

 corner points $z=16x+20y$ $A(10,0)$ 160 $B(2,4)$ 112 minimum $C(1,5)$ 116 $D(0,8)$ 160

The feasible region is unbounded , therefore 112 may or may not be minimum value of Z .

For this we draw $16x+20y< 112$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $16x+20y< 112$ .

Hence , Z has minimum value 112 at point $B(2,4)$

Question:4 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:

 Types of toys Machines I II III A 12 18 6 B 6 0 9

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.

Let x and y toys of type A and type B.

Mathematical formulation of given problem is as follows:

Minimize : $z=7.5x+5y$

Subject to constraint ,

$2x+y\leq 60$

$x\leq 20$

$2x+3y \leq 120$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(20,0),B(20,20),C(15,30),D(0,40)$

The value of Z at corner points is as shown :

 corner points $z=7.5x+5y$ $A(20,0)$ 150 $B(20,20)$ 250 $C (15,30)$ 262.5 maximum $D(0,40)$ 200

Therefore 262.5 may or may not be maximum value of Z .

Hence , Z has maximum value 262.5 at point $C (15,30)$

Question:5 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?

Let airline sell x tickets of executive class and y tickets of economy class.

Mathematical formulation of given problem is as follows:

Minimize : $z=1000x+600y$

Subject to constraint ,

$x+y\leq 200$

$x\geq 20$

$y-4x\geq 0$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(20,80),B(40,160),C(20,180)$

The value of Z at corner points is as shown :

 corner points $z=1000x+600y$ $A(20,80)$ 68000 $B(40,160)$ 136000 maximum $C (20,180)$ 128000

therefore 136000 is maximum value of Z .

Hence , Z has maximum value 136000 at point $B(40,160)$

Question:6 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:

 Transportation cost per quintal (in Rs) From/To A B D 6 4 E 3 2 F 2.50 3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Let godown A supply x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be supplied to shop F. Requirements at shop D is 60 since godown A supply x .Therefore remaining (60-x) quintals of grain will be transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:

$x,y\geq 0$ and $100-x-y\geq 0$

$x,y\geq 0$ and $x+y\leq 100$

$60-x\geq 0,50-y\geq 0\, \, \, and\, \, x+y-60\geq 0$

$\Rightarrow \, \, \, \, x\leq 60,y\leq 50,x+y\geq 60$

Total transportation cost z is given by ,

$z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)$

$z=2.5x+1.5y+410$

Mathematical formulation of given problem is as follows:

Minimize : $z=2.5x+1.5y+410$

Subject to constraint ,

$x+y\leq 100$

$x\leq 60$

$y\leq 50$

$x+y\geq 60$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(60,0),B(60,40),C(50,50),D(10,50)$

The value of Z at corner points is as shown :

 corner points $z=2.5x+1.5y+410$ $A(60,0)$ 560 $B(60,40)$ 620 $C(50,50)$ 610 $D(10,50)$ 510 minimum

therefore 510 may or may not be minimum value of Z .

Hence , Z has miniimum value 510 at point $D(10,50)$

Question:7 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:

 Distance in (km.) From/To A B D 7 3 E 6 4 F 3 2

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?

Let x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be supplied from A to petrol pump F.

Requirements at petrol pump D is 4500 L. since x L A are transported from depot A,remaining 4500-x L will be transported from petrol pump B

Similarly, (3000-y)L and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot B to petrol E and F respectively.

The problem can be represented diagrammatically as follows:

$x,y\geq 0$ and $7000-x-y\geq 0$

$x,y\geq 0$ and $x+y\leq 7000$

$4500-x\geq 0,3000-y\geq 0\, \, \, and\, \, x+y-3500\geq 0$

$\Rightarrow \, \, \, \, x\leq 4500,y\leq 3000,x+y\geq 3500$

Cost of transporting 10 L petrol =Re 1

Cost of transporting 1 L petrol $=\frac{1}{10}$

Total transportation cost z is given by ,

$z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$

$z=0.3x+0.1y+3950$

Mathematical formulation of given problem is as follows:

Minimize : $z=0.3x+0.1y+3950$

Subject to constraint ,

$x+y\leq 7000$

$x\leq 4500$

$y\leq 3000$

$x+y\geq 3500$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(3500,0),B(4500,0),C(4500,2500),D(4000,3000),E(500,3000)$

The value of Z at corner points is as shown :

 corner points $z=0.3x+0.1y+3950$ $A(3500,0)$ 5000 $B(4500,0)$ 5300 $C(4500,2500)$ 5550 $E(500,3000)$ 4400 minimum $D(4000,3000)$ 5450

Hence , Z has miniimum value 4400 at point $E(500,3000)$

Question:8 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?

 Kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Let fruit grower use x bags of brand P and y bags of brand Q.

Mathematical formulation of given problem is as follows:

Minimize : $z=3x+3.5y$

Subject to constraint ,

$x+2y\geq 240$

$x+0.5y\geq 90$

$1.5x+2y\geq 310$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(140,50),C(40,100),B(20,140)$

The value of Z at corner points is as shown :

 corner points $z=3x+3.5y$ $A(140,50)$ 595 $B(20,140)$ 550 $C(40,100)$ 470 minimum

Therefore 470 is minimum value of Z .

Hence , Z has minimum value 470 at point $C(40,100)$

Question:9 Reference of Que 8 : 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 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?

 Kg per bag Brand A Brand P Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Let fruit grower use x bags of brand P and y bags of brand Q.

Mathematical formulation of given problem is as follows:

Maximize : $z=3x+3.5y$

Subject to constraint ,

$x+2y\geq 240$

$x+0.5y\geq 90$

$1.5x+2y\geq 310$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $B(20,140),A(140,50),C(40,100)$

The value of Z at corner points is as shown :

 corner points $z=3x+3.5y$ $A(140,50)$ 595 maximum $B(20,140)$ 550 $C(40,100)$ 470 minimum

therefore 595 is maximum value of Z .

Hence , Z has minimum value 595 at point $A(140,50)$

Question:10 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?

Let x and y be number of dolls of type A abd B respectively that are produced per week.

Mathematical formulation of given problem is as follows:

Maximize : $z=12x+16y$

Subject to constraint ,

$x+y\leq 1200$

$y\leq \frac{x}{2}\Rightarrow x\geq 2y$

$x-3y\leq 600$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(600,0),B(1050,150),C(800,400)$

The value of Z at corner points is as shown :

 corner points $z=12x+16y$ $A(600,0)$ 7200 $B(1050,150)$ 15000 $C(800,400)$ 16000 Maximum

Therefore 16000 is maximum value of Z .

Hence , Z has minimum value 16000 at point $C(800,400)$

## More About NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise:

There are 10 questions in the miscellaneous exercise chapter 12 Class 12. Solving all these questions gives a good knowledge about the NCERT Class 12th chapter linear programming. Students have to solve the NCERT syllabus exercises and solved examples in order to get a good idea of topics discussed in the chapter and to get a good score in the final exam.

Also Read| Linear Programming Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise.

• Important questions to understand the concepts explained in the chapter are given in Class 12 Maths chapter 12 miscellaneous exercise solutions.

• All questions of miscellaneous exercise chapter 12 Class 12 are important and will be helpful in the board exam.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 12

• NCERT Solutions for Class 12 Maths Chapter 12

## NCERT Solutions Subject Wise

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• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology