NCERT Solutions for Exercise 12.1 Class 12 Maths Chapter 12 - Linear Programming

NCERT solutions for exercise 12.1 Class 12 Maths chapter 12 gives practice questions to understand linear programming problems. There are 10 questions explained in exercise 12.1 Class 12 Maths. There are many methods to solve linear programming problems. Here in this Class 12 NCERT Mathematics chapter, only graphical methods for solving linear programming problems are discussed. In the NCERT Solutions for class 12 maths chapter 12 exercise 12.1 all the 10 questions are solved using graphs. Class 12 Maths chapter 12 exercise 12.1 solves problems related to maximising or minimising linear functions subjected to certain constraints. These constraints are a set of linear inequalities. Concepts of linear inequalities are already introduced in Class 11 Mathematics NCERT book. Other than Class 12 Maths chapter 12 exercise 12.1 solutions there are two more exercises.

• Linear Programming Exercise 12.2

• Linear Programming Miscellaneous Exercise

Linear Programming Class 12 Chapter12-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.

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 $D(0,4)$

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

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.

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

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.

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.

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.

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.

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 Maximise $z=x+2y$ Subject to $x+2y\geq 100,2x-y\leq 0,2x+y\leq 200,x,y,\geq 0$ Show that the minimum of Z occurs at more than two points.

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 $D(0,200)$ .

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.

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 to $x-y\leq -1,-x+ y\leq 0,x,y,\geq 0.$ Show that the minimum of Z occurs at more than two points.

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.

More About NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1

In order to understand the concepts well it is important to solve the NCERT syllabus exercise questions. The NCERT solutions for Class 12 Maths chapter 12 exercise 12.1 helps in solving the first exercise of the chapter linear programming. Students can make use of Class 12 Maths chapter 12 exercise 12.1 solutions for preparation of board exams as well as engineering entrance exam like JEE Mains.

Also Read| Linear Programming Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1

• Class 12th Maths chapter 12 exercise 12.1 are prepared by the best faculties of Mathematics

• All main topics are covered and exercise 12.1 Class 12 Maths gives answers to all the questions and are in detail

• Students can use Class 12 Maths chapter 12 exercise 12.1 to prepare for CBSE exams

Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 12

• NCERT Solutions for Class 12 Maths Chapter 12

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