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: 12.1

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

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 $D(0,4)$

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 $B(4,0)$

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

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 $D(\frac{20}{19},\frac{45}{19})$

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 $B(\frac{3}{2},\frac{1}{2})$

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 $B(4,3)$

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

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  $Z = x + 2y$.

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 $D(120,0)$

The region determined by constraints $x+2y\geq 100,2x-y\leq 0,2x+y\leq 200,x,y,\geq 0$is 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 $D(0,200)$.

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.

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

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

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:

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   $B(2,\frac{1}{2})$.

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).

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

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 .

(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.

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.

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.

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.

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.

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.

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.

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.

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  $,B(100,80)$

$2x+y \leq 10,x+3y \leq 15,x,y\geq 0$ 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

$(A) p=q$

$(B)p=2q$

$(C)p=3q$

$(D)q=3p$

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

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.

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 $C(3,6)$.

 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  $B(2,4)$

 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  $C (15,30)$

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  $B(40,160)$

 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  $D(10,50)$

 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  $E(500,3000)$

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  $C(40,100)$

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  $A(140,50)$

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

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