NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

 

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming: We often come across problems where we seek to maximize profit or minimize cost. NCERT solutions for class 12 maths chapter 12 linear programming is helpful to find the most optimal solution for such problems. In the solutions of NCERT for class 12 maths chapter 12 linear programming you will learn to formulate these real-life problems into a mathematical model. In class 11th you have already learnt linear inequalities and their solutions by the graphical method and systems of linear equations and their applications in daily life problems. In this chapter, you are going to deal with problems on linear programming like maximization and minimization of equations, mathematical and graphical methods to solve problems of linear programming. In solutions of NCERT for class 12 maths chapter 12 linear programming article, questions from all these topics are covered. Check all NCERT solutions from class 6 to 12 at a single place, which will help you get a better understanding of concepts in a much easy way.

Generally, one question( 5 marks) from this chapter is asked in the 12th board final examination. In this chapter, there are 2 exercises with 21 questions. In this article, you will find NCERT solutions for class 12 maths chapter 12 linear programming which are prepared and explained in a detailed manner. It will help you to score well in the board exam as well as in the competitive exams. 

Let's take an NCERT problem - A furniture dealer deals in only two items–chairs and tables. Has storage space of at most 60 pieces and He has Rs 50,000 to invest. A chair costs Rs 500 and A table costs Rs 2500. He estimates that from the sale of one chair, he can make a profit of Rs 75 and that from the sale of one table a profit of Rs 250. He wants to know how many chairs and tables he should buy from the available money so as to maximize his total profit, assuming that he can sell all the items which he buys.

That type of problem which involves- minimize profit or maximize cost is called optimization problems. Linear programming problems are a very important class of optimization problems. The above-stated problem is an example of linear programming. This chapter is an important chapter because of its wide applicability in industry, management science, commerce, etc. In this chapter, you will learn mathematical and graphical methods to solve problems of linear programming.

Topics of NCERT class 12 maths chapter-12 Linear Programming

12.1 Introduction

12.2 Linear Programming Problem and its Mathematical Formulation

12.2.1 Mathematical formulation of the problem

12.2.2 Graphical method of solving linear programming problems

12.3 Different Types of Linear Programming Problems

NCERT solutions for class 12 maths chapter 12 Linear Programming- Exercise Questions

NCERT solutions for class 12 maths chapter 12 linear programming-Exercise: 12.1

Question:1 Solve the following Linear Programming Problems graphically: Maximise  Z = 3x + 4y Subject to the constraints  x+y\leq 4,x\geq 0,y\geq 0. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, x+y\leq 4,x\geq 0,y\geq 0. is as follows,

 Chapter 12 Linear Programming Question 1

The region A0B represents the feasible region

The corner points of the feasible region are B(4,0),C(0,0),D(0,4)

Maximize  Z = 3x + 4y

 The value of these points at these corner points are : 

Corner points 

          Z = 3x + 4y

 

        B(4,0)

              12

 

       C(0,0)

             0

 

        D(0,4)

               16

maximum

The maximum value of Z is 16  at D(0,4)

Question:2 Solve the following Linear Programming Problems graphically: Minimise  z=-3x+4y Subject to .x+2y\leq 8,3x+2y\leq 12,x\geq 0,y\geq 0. Show that the minimum of Z occurs at more than two points

Answer:

The region determined by constraints, x+2y\leq 8,3x+2y\leq 12,x\geq 0,y\geq 0. is as follows,

               

The corner points of feasible region are A(2,3),B(4,0),C(0,0),D(0,4)

 The value of these points at these corner points are : 

Corner points 

          z=-3x+4y

 

        A(2,3)

               6

 

        B(4,0)

              -12

Minimum

       C(0,0)

             0

 

        D(0,4)

               16

 

The minimum value of Z is -12  at B(4,0)

Question:3 Solve the following Linear Programming Problems graphically: Maximise  Z = 5x + 3y Subject to 3x + 5y \leq 15,5x+2y\leq 10, x\geq 0,y\geq 0 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints,  3x + 5y \leq 15,5x+2y\leq 10, x\geq 0,y\geq 0 is as follows :

          Chapter 12  Linear Programming  Question 3

The corner points of feasible region are A(0,3),B(0,0),C(2,0),D(\frac{20}{19},\frac{45}{19})

 The value of these points at these corner points are : 

Corner points 

          Z = 5x + 3y

 

        A(0,3)

               9

 

        B(0,0)

              0

 

       C(2,0)

              10

 

        D(\frac{20}{19},\frac{45}{19})  

               \frac{235}{19}

 Maximum 

The maximum value of Z is  \frac{235}{19}at D(\frac{20}{19},\frac{45}{19})

Question:4 Solve the following Linear Programming Problems graphically: Minimise  Z = 3x + 5y Such that  x+3y\geq 3,x+y\geq 2,x,y\geq 0. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x+3y\geq 3,x+y\geq 2,x,y\geq 0.is as follows,

               

The feasible region is unbounded as shown.

The corner points of the feasible region are A(3,0),B(\frac{3}{2},\frac{1}{2}),C(0,2)

 The value of these points at these corner points are : 

Corner points 

          Z = 3x + 5y

 

        A(3,0)

               9

 

        B(\frac{3}{2},\frac{1}{2})     

              7

Minimum

       C(0,2)

              10

 

        

               

 

  The feasible region is unbounded, therefore 7 may or may not be the minimum value of Z .

For this, we draw 3x + 5y< 7 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.Z = 3x + 5y

Hence, Z has a minimum value of 7 at B(\frac{3}{2},\frac{1}{2})

Question:5 Solve the following Linear Programming Problems graphically: Maximise Z = 3x + 2y Subject to x+2y\leq 10,3x+y\leq 15,x,y\geq 0      Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, x+2y\leq 10,3x+y\leq 15,x,y\geq 0  is as follows,

               

The corner points of feasible region are A(5,0),B(4,3),C(0,5)

 The value of these points at these corner points are : 

Corner points 

          Z = 3x + 2y

 

        A(5,0)

               15

 

        B(4,3)

               18

Maximum

       C(0,5)

                10

 

        

                

 

The maximum  value of Z is  18  at B(4,3)

Question:6 Solve the following Linear Programming Problems graphically: Minimise Z = x + 2y Subject to 2x+y\geq 3,x+2y\geq 6,x,y\geq 0.

 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints 2x+y\geq 3,x+2y\geq 6,x,y\geq 0.is as follows,

               

The corner points of the feasible region are A(6,0),B(0,3)

 The value of these points at these corner points are : 

Corner points 

          Z = x + 2y

        A(6,0)

               6

   B(0,3)

              6

Value of Z is the same at both points.A(6,0),B(0,3)

If we take any other point like (2,2) on line Z = x + 2y , then Z=6.

Thus the minimum value of Z occurs at more than 2 points .

Therefore, the value of Z is minimum at every point on the line  Z = x + 2y.

Question:7 Solve the following Linear Programming Problems graphically: Minimise and Maximise z=5x+10y Subject to x+2y\leq 120,x+y\geq 60,x-2y\geq 0,x,y\geq 0 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, x+2y\leq 120,x+y\geq 60,x-2y\geq 0,x,y\geq 0 is as follows,

               

The corner points of feasible region are A(40,20),B(60,30),C(60,0),D(120,0)

 The value of these points at these corner points are : 

Corner points 

          z=5x+10y

 

        A(40,20)

               400

 

        B(60,30)

               600

Maximum

       C(60,0)

               300

Minimum

        D(120,0)

               600

maximum

The minimum value of Z is 300  at  C(60,0)  and maximum value is 600 at all points joing line segment  B(60,30) and D(120,0)

Question:8 Solve the following Linear Programming Problems graphically: Minimise and Maximisez=x+2y  Subject tox+2y\geq 100,2x-y\leq 0,2x+y\leq 200,x,y,\geq 0 Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x+2y\geq 100,2x-y\leq 0,2x+y\leq 200,x,y,\geq 0is as follows,

               

The corner points of the feasible region are A(0,50),B(20,40),C(50,100),D(0,200)

 The value of these points at these corner points are : 

Corner points 

          z=x+2y

 

        A(0,50)

               100

Minimum

        B(20,40)

              100

Minimum

       C(50,100)

              250

 

        D(0,200)

               400

Maximum

The minimum value of Z is 100  at all points on the line segment joining points  A(0,50) and B(20,40).

The maximum value of Z is 400 at D(0,200).

Question:9 Solve the following Linear Programming Problems graphically: Maximise Z = -x+2y Subject to the constraints:x\geq 3,x+y\geq 5,x+2y\geq 6,y\geq 0. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x\geq 3,x+y\geq 5,x+2y\geq 6,y\geq 0.is as follows,

               

The corner points of the feasible region are A(6,0),B(4,1),C(3,2)

 The value of these points at these corner points are : 

Corner points 

          Z = -x+2y

 

        A(6,0)

              - 6

minimum

        B(4,1)

              -2

 

       C(3,2)

                1

maximum 

        

               

 

The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.

For this, we draw -x+2y> 1 and check whether resulting half plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with a feasible region.

Hence , Z =1 is  not maximum value , Z has no maximum value. 

Question:10 Solve the following Linear Programming Problems graphically: Maximise Z = x + y, Subject tox-y\leq -1,-x+ y\leq 0,x,y,\geq 0. Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints x-y\leq -1,-x+ y\leq 0,x,y,\geq 0.is as follows,

   Chapter 12 � Linear Programming Question 10

There is no feasible region and thus, Z has no maximum value.

CBSE NCERT solutions for class 12 maths chapter 12 linear programming-Exercise: 12.2

Question:1 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 Rs 60/kg and Food Q costs Rs.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.

Answer:

Let mixture contain x kg of food P and y kg of food 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 mixture must contain 8 units of Vitamin A and 11 units of Vitamin B.

Therefore, we have 

               3x+4y\geq 8

               5x+2y\geq 11

Total cost is Z. Z=60x+80y

Subject to constraint,

               3x+4y\geq 8

               5x+2y\geq 11

                x\geq 0,y\geq 0

The feasible  region determined by constraints is as follows:

    Chapter 12 � Linear Programming

It can be seen that a feasible region is unbounded.

The corner points of the feasible region are A(\frac{8}{3},0),B(2,\frac{1}{2}),C(0,\frac{11}{2})

The value of Z at corner points is as shown :

 corner points 

Z=60x+80y

 

   A(\frac{8}{3},0)

            160

MINIMUM

B(2,\frac{1}{2})

           160

minimum

C(0,\frac{11}{2})

            440

 

Feasible region is unbounded, therefore 160 may or may not be the minimum value of Z.

For this, we draw 60x+80y< 160\, \, or \, \, \, 3x+4y< 8 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.\, \, 3x+4y< 8

Hence, Z has a minimum value 160  at line segment joining points  A(\frac{8}{3},0)     and   B(2,\frac{1}{2}).

 

Question:2 One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Answer:

Let there be x cakes of first kind and y cakes of the second kind.Thus, x\geq 0,y\geq 0.

The given information can be represented in the table as :

 

Flour(g)

fat(g)

Cake of kind  x 

200

25

Cake of kind y

100

50

Availability

5000

1000

Therefore,

               200x+100y\leq 5000

               \Rightarrow \, \, \, \, 2x+y\leq 50

                    . \, \, 25x+50y\leq 10000

                         \Rightarrow \, \, x+2y\leq 400

The total number of cakes, Z.   Z=X+Y

Subject to constraint,

   \Rightarrow \, \, \, \, 2x+y\leq 50 

   \Rightarrow \, \, x+2y\leq 400

          x\geq 0,y\geq 0

The feasible  region determined by constraints is as follows:

    

The corner points of the feasible region are A(25,0),B(20,10),C(0,20),D(0,0)

The value of Z at corner points is as shown :

 corner points 

Z=X+Y

 

  A(25,0)

          25

 

   B(20,10)

          30

maximum

C(0,20)

D(0,0)

      20       

       0

 

minimum

The maximum cake can be made 30 (20 of the first kind and 10 of the second kind).

 

Question:3 A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

 (i)   What number of rackets and bats must be made if the factory is to work at full capacity? 

Answer:

Let number of rackets be x and number of bats be y.

the  machine time availability is not more than 42 hours.

i.e.          1.5x+3y\leq 42

 craftsman’s time availability is 24 hours

i.e.        3x+y\leq 24

 The factory has to work at full capacity.

Hence, 1.5x+3y= 42...............1

                  3x+y= 24...............2

Solving equation 1 and 2, we have 

                   x=4\, \, and\, \, \, y=12

Thus, 4 rackets and 12 bats are to be made .

Question:3 A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(ii)  If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

Answer:

Let the number of rackets is x and the number of bats is y.

the machine time availability is not more than 42 hours.

 craftsman’s time availability is 24 hours

The given information can be repreented in table as shown :

 

racket

bat  

availability

machine time 

1.5

3

42

craftman's time

3

1

24

  1.5x+3y\leq 42

  3x+y\leq 24

        x,y\geq 0

The profit on the bat is 10 and on the racket is 20.

              Z=20x+10y

The mathematical formulation is :

maximise Z=20x+10y

subject to constraints,

  1.5x+3y\leq 42

  3x+y\leq 24

        x,y\geq 0

The feasible  region determined by constraints is as follows:

The corner points are A(8,0),B(4,12),C(0,14),D(0,0)

The value of Z at corner points is as shown :

CORNER POINTS

Z=20x+10y

 

A(8,0)

             160

 

B(4,12)

               200

maximum 

C(0,14)

                140

 

D(0,0)

                  0

 

Thus, the maximum profit of the factory when it works at full capacity is 200.

Question:4  A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

Answer:

Let packages of nuts be x and packages of bolts be y  .Thus, x\geq 0,y\geq 0.

The given information can be represented in table as :

 

bolts 

nuts  

availability

machine A

  1

    3

    12

machine B

   3

     1

     12

 

 

 

 

 Profit on a package of nuts is Rs. 17.5  and on package of bolt is 7.

Therefore, constraint  are  

                x+3y\leq 12

               3x+y\leq 12

                x\geq 0,y\geq 0

                 Z= 17.5x+7y

The feasible  region determined by constraints is as follows:

    

 

The corner points of feasible region are  A(4,0),B(3,3),C(0,4),D(0,0)

The value of Z at corner points is as shown :

             

Corner points

Z= 17.5x+7y

 

A(4,0)

                   70

 

B(3,3)

                   73.5

maximum

C(0,4)

                    28

 

D(0,0)

                    0

 

 The maximum value of z is 73.5 at B(3,3).

Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.

Question:5 A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Answer:

Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, x\geq 0,y\geq 0.

The given information can be represented in the table as :

 

screw A  

screw B  

availability

Automatic machine

  4

    6

    4\times 60=240

hand operated machine 

   6

     3

     4\times 60=240

 

 

 

 

 Profit on a package of screw A  is Rs.7 and on the package of screw B  is 10.

Therefore, the constraint is  

                4x+6y\leq 240

               6x+3y\leq 240

                x\geq 0,y\geq 0

                 Z= 7x+10y

The feasible  region determined by constraints is as follows:

    

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

The value of Z at corner points is as shown :

Corner points

Z= 7x+10y

 

A(40,0)

                   280

 

B(30,20)

                   410

maximum

C(0,40)

                    400

 

D(0,0)

                    0

 

 The maximum value of z is 410 at B(30,20).

Thus, 30 packages of screw A  and 20 packages of screw B should be manufactured every day to get maximum profit.

Question:6 A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?

Answer:

Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, x\geq 0,y\geq 0.

The given information can be represented in the table as :

 

lamps 

shades   

availability

machine (h)

  2

    1

    12

sprayer (h)

   3

     2

     20

 

 

 

 

 Profit on a lamp is Rs. 5  and on the shade is 3.

Therefore, constraint is  

                2x+y\leq 12

               3x+2y\leq 20

                x\geq 0,y\geq 0

                 Z= 5x+3y

The feasible  region determined by constraints is as follows:

       

The corner points of the feasible region are  A(6,0),B(4,4),C(0,10),D(0,0)

The value of Z at corner points is as shown :

Corner points

Z= 5x+3y

 

A(6,0)

               30

 

B(4,4)

                32

maximum

C(0,10)

                 30

 

D(0,0)

                 0

 

 The maximum value of z is 32 at B(4,4).

Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.

Question:7 A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Answer:

Let  x be Souvenirs of type A and  y be Souvenirs of type B  .Thus, x\geq 0,y\geq 0.

The given information can be represented in table as :

 

Type A 

Type B  

availability

cutting 

  5

    8

    (3\times 60)+20=200

asembling

   10

     8

     4\times 60=240

 

 

 

 

 Profit on type A Souvenirs is Rs. 5  and on type B Souvenirs  is 6.

Therefore, constraint  are  

                5x+8y\leq 200

               10x+8y\leq 240

                x\geq 0,y\geq 0

                 Z=5x+6y

The feasible  region determined by constraints is as follows:

    

The corner points of feasible region are  A(24,0),B(8,20),C(0,25),D(0,0)

The value of Z at corner points is as shown :

Corner points

Z=5x+6y

 

A(24,0)

            120

 

B(8,20)

             160

maximum

C(0,25)

              150

 

D(0,0)

              0

 

 The maximum value of z is 160 at B(8,20) .

Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.

Question:8 A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Answer:

Let  merchant plans has  personal computers  x  desktop model and y portable model

 .Thus, x\geq 0,y\geq 0.

The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.

Merchant can invest  Rs 70 lakhs maximum.

   25000x+40000y\leq 7000000

  5x+8y\leq 1400

the total monthly demand of computers will not exceed 250 units.

       x+y\leq 250

profit on the desktop model  is Rs 4500 and on portable model is Rs 5000.

Total profit = Z , Z=4500x+5000y

The mathematical formulation of given problem is :
                5x+8y\leq 1400

               x+y\leq 250

                x\geq 0,y\geq 0

                 Z=4500x+5000y

The feasible  region determined by constraints is as follows:

    

The corner points of feasible region are  A(250,0),B(200,50),C(0,175),D(0,0)

The value of Z at corner points is as shown :

Corner points

Z=4500x+5000y

 

A(250,0)

                   1125000

 

B(200,50)

                   1150000

maximum

C(0,175)

                  875000  

 

D(0,0)

                    0

 

 The maximum value of z is 1150000 at B(200,50).

Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

Question:9 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 Rs 4 per unit food and F2 costs Rs 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. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

Answer:

Let  diet contain x unit of food F1  and y unit of foof F2  .Thus, x\geq 0,y\geq 0.

The given information can be represented in table as :

 

Vitamin 

minerals 

cost per unit 

foof F1

  3

    4

      4

food F2

   6

     3

       6

 

    80

    100

 

 Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint  are  

                3x+4y\geq 4

               6x+3y\geq 6

                x\geq 0,y\geq 0

                 Z= 4x+6y

The feasible  region determined by constraints is as follows:

    

We can see feseable region is unbounded.

The corner points of feasible region are  A(\frac{80}{3},0),B(24,\frac{4}{3}),C(0,\frac{100}{3})

The value of Z at corner points is as shown :

Corner points

Z= 4x+6y

 

A(\frac{80}{3},0)

            106.67

 

B(24,\frac{4}{3}),

              104

minimum

C(0,\frac{100}{3})

                200

maximum

 

                    

 

  Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw 4x+6y< 104   or   2x+3y< 52  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  2x+3y< 52.

Hence , Z has minimum value 104.

Question:10 There are two types of fertilisers F1 and F2 . F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer:

Let farmer buy x kg of fertilizer F1 and y kg of F2  .Thus, x\geq 0,y\geq 0.

The given information can be represented in table as :

 

Nitrogen

phosphoric acid 

Cost 

F1

  10

    6

    6

F2

   5

     10

     5

 requirement 

   14

    14

 

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen 

               10\%x+5\%y\geq 14

                   \frac{x}{10}+\frac{y}{20}\geq 14

               2x+y\geq 280

F1 contain 6% phophoric acid  and F2 contain 10% phosphoric acid  .Farmer requires atleast 14 kg of nitrogen 

               6\%x+10\%y\geq 14

                   \frac{6x}{100}+\frac{y}{20}\geq 14

               3x+56y\geq 700

Total cost is  Z . Z=6x+5y

Subject to constraint ,

               2x+y\geq 280

               3x+56y\geq 700

                x\geq 0,y\geq 0

             Z=6x+5y

The feasible  region determined by constraints is as follows:

    

It can be seen that feasible region is unbounded.

The corner points of feasible region are A(\frac{700}{3},0),B(100,80),C(0,280)

The value of Z at corner points is as shown :

 corner points 

Z=6x+5y

 

   A(\frac{700}{3},0)

            1400

 

,B(100,80)

           1000

minimum

C(0,280)

            1400

 

  Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw 6x+5y< 1000 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  6x+5y< 1000.

Hence , Z has minimum value 1000  at  point  ,B(100,80)

Question:11 The corner points of the feasible region determined by the following system of linear inequalities:

        2x+y \leq 10,x+3y \leq 15,x,y\geq 0 are (0,0),(5,0),(3,4) and (0,5). Let Z=px+qy,                where p,q > 0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is

               (A) p=q

                (B)p=2q

                (C)p=3q

                (D)q=3p

Answer:

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points (3,4)\, \, and\, \, \, (0,5).

\therefore Value of Z at (3,4) =value of Z at (0,5)

   \Rightarrow \, \, \, p(3)+q(4)=p(0)+q(5)

\Rightarrow \, \, \, 3p+4q=5q

\Rightarrow \, \, \, q=3p

Hence, D is correct option.

NCERT solutions for class 12 maths chapter 12 linear programming-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?

Answer:

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 pointB(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?

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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

 

Answer:

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

 

Answer:

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?

Answer:

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) 

NCERT solutions for class 12 maths chapter wise

chapter 1

Solutions of NCERT for class 12 maths chapter 1 Relations and Functions

chapter 2

CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

chapter 3

NCERT solutions for class 12 maths chapter 3 Matrices

chapter 4

Solutions of NCERT for class 12 maths chapter 4 Determinants

chapter 5

CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

chapter 6

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

chapter 7

Solutions of NCERT for class 12 maths chapter 7 Integrals

chapter 8

CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals

chapter 9

NCERT solutions for class 12 maths chapter 9 Differential Equations

chapter 10

Solutions of NCERT for class 12 maths chapter 10 Vector Algebra

chapter 11

CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

chapter 12

NCERT solutions for class 12 maths chapter 12 Linear Programming

chapter 13

Solutions of NCERT for class 12 maths chapter 13 Probability

NCERT solutions for class 12 subject wise

Benefits of NCERT solutions

  • NCERT solutions are prepared and explained in a step-by-step manner, so it is very easy for you to understand the concepts.

  • These NCERT solutions will give you new ways of solving the problems.

  • NCERT solutions for class 12 maths chapter 12 linear programming will help you to score good marks in the exam as these questions are answered by the experts who know how best to answer the questions in the board exam. 

  • Miscellaneous exercise is very important if you wish to develop a grip on the concepts. In solutions of NCERT for class 12 maths chapter 12 linear programming, you will get solutions for miscellaneous exercise too.  

Happy learning !!!

 

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