NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

Class 12 Maths Chapter 12, Linear Programming, explains how mathematical methods can be used to solve optimisation problems based on real-life situations. Students learn how to formulate a Linear Programming Problem (LPP), write the objective function and constraints, represent linear inequalities on a graph, identify the feasible region, and find the maximum or minimum value of the objective function. The chapter also covers applications of linear programming in areas such as business and economics.

NCERT Solutions for Class 12 Maths Chapter 12 by ALLEN follow the latest CBSE syllabus and cover the questions from the NCERT textbook. Step-by-step solutions help students understand linear programming problems, graphical representation, feasible regions, constraints, objective functions, and the corner point method. Regular practice can help students solve LPP questions accurately and prepare for CBSE board exams and competitive examinations.

1.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Class 12 Maths Chapter 12, Linear Programming, focuses on solving optimisation problems using graphical methods. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 12 include:

Linear Programming Problem (LPP): Learn how real-life situations can be expressed as mathematical optimisation problems.
Objective Function: Understand how to define a function that needs to be maximised or minimised based on the given problem.
Constraints: Study the linear inequalities that represent restrictions and limit the possible solutions.
Feasible Region: Identify the common region on a graph that satisfies all the given constraints.
Corner Point Method: Find the optimal solution by evaluating the objective function at the corner points of the feasible region.
Graphical Representation: Learn how to draw and interpret graphs of linear inequalities to identify the feasible region and solve LPP questions.

2.0NCERT Class 12 Maths Chapter 12 Linear Programming : Detailed Solutions

EXERCISE - 12.1

Solve the following Linear Programming problems graphically.

Maximize $Z = 3 x + 4 y$ , Subject to the constraints

$x + y \leq 4, x \geq 0 and y \geq 0.$

Sol. We have to Maximize $Z = 3 x + 4 y$ Subject to constraints $x + y \leq 4, x \geq 0, y \geq 0$ Firstly, draw the graph of the line $x + y = 4$ Then, putting $(0, 0)$ in the inequality $x + y \leq 4$ we have $0 + 0 \leq 4 \Rightarrow\leq 4$ (Which is true) So, the half plane is towards the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant.

∴ Feasible region is OABO. The corner points of the feasible region are $O (0, 0), A (4, 0)$ and $B (0, 4)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 4y$
O(0,0)	0
A(4,0)	12
B(0,4)	$16 \to$ Maximum

Therefore, the maximum value of Z is 16 at the point B(0,4).

Minimize $Z = - 3 x + 4 y$ , subject to constraints

$x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0 and y \geq 0 .$

Sol. We have to Minimize $Z = - 3 x + 4 y$ Subject to constraints

$x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x + 2 y = 8$ Putting $(0, 0)$ in the inequality $x + 2 y \leq 8$ , we have $0 + 0 \leq 8 \Rightarrow 0 \leq 8$ (Which is true) So, the half plane is towards the origin.

Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $3 x + 2 y = 12$

x	0	4
y	6	0

Putting $(0, 0)$ in the inequality $3 x + 2 y \leq 12$ we have $3 \times 0 + 2 \times 0 \leq 12$ $\Rightarrow 0 \leq 12$ (Which is true) So, the half plane is towards the origin. ∴ Feasible region is OABCO. On solving equations

$x + 2 y = 8 and 3 x + 2 y = 12,$

we get $x = 2$ and $y = 3$ ∴ Intersection point B is $(2, 3)$ The corner points of the feasible region are $O (0, 0), A (4, 0), B (2, 3)$ and $C (0, 4)$ .

The values of Z at these points are as follows.

Corner Points	$Z = - 3x + 4y$
$O (0, 0)$	0
A(4,0)	-12 →Minimum
B(2,3)	6
C(0,4)	16

Therefore, the minimum value of Z is -12 at the point A(4,0).

Maximize $Z = 5 x + 3 y$ , subject to constraints

$3 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0 and y \geq 0.$

Sol. We have to Maximize $Z = 5 x + 3 y$ Subject to constraints

$3 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0, y \geq 0$

Firstly, draw the graph of the line,

$3 x + 5 y = 15$

Putting $(0, 0)$ in the inequality $3 x + 5 y \leq 15$ , we have

$3 \times 0 + 5 \times 0 \leq 15$

$\Rightarrow 0 \leq 15$ (Which is true) So, the half plane is towards the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $5 x + 2 y \leq 10$ Putting $(0, 0)$ in the inequality $5 x + 2 y \leq 10$ we have $5 \times 0 + 2 \times 0 \leq 10$ $\Rightarrow 0 \leq 10$ (Which is true) So, the half plane is towards the origin. On solving equations

$3 x + 5 y = 15 and 5 x + 2 y = 10$

we get $x = \frac{20}{19}$ and $y = \frac{45}{19}$

Coordinates of point B is $(\frac{20}{19}, \frac{45}{19})$

∴ Feasible region is OABCO The corner points of the feasible region are $O (0, 0), A (2, 0), B (\frac{20}{19}, \frac{45}{19})$ and $C (0, 3)$ . The values of Z at these points are as follows:

Corner Points	$Z = 5x + 3y$
$O (0, 0)$	0
A(2,0)	10
$B (\frac{20}{19}, \frac{45}{19})$	$\frac{235}{19} \to$ Maximum
C(0,3)	9

Therefore, the maximum value of Z is $\frac{235}{19}$ at the point $B (\frac{20}{19}, \frac{45}{19})$ .

Minimize $Z = 3 x + 5 y$ subject to constraints

$x + 3 y \geq 3, x + y \geq 2 and x, y \geq 0.$

Sol. We have to Minimize $Z = 3 x + 5 y$ Subject to constraints

$x + 3 y \geq 3, x + y \geq 2, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x + 3 y = 3$ Putting $(0, 0)$ in the inequality $x + 3 y \geq 3$ , we have $0 + 3 \times 0 \geq 3 \Rightarrow 0 \geq 3$ (Which is false)

So, the half plane is away from the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $x + y = 2$ Putting $(0, 0)$ in the inequality $x + y \geq 2$ we have $0 + 0 \geq 2 \Rightarrow 0 \geq 2$ (Which is false) So, the half plane is away from the origin. It can be seen that the feasible region is unbounded. On solving equations $x + y = 2$ and $x + 3 y = 3$ , we get $x = \frac{3}{2}$ and $y = \frac{1}{2}$ ∴ Intersection point is $B (\frac{3}{2}, \frac{1}{2})$

The corner points of the feasible region are $A (3, 0), B (\frac{3}{2}, \frac{1}{2})$ and $C (0, 2)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 5y$
A(3,0)	9
B $(\frac{3}{2}, \frac{1}{2})$	$7 \to$ Minimum
C(0,2)	10

As the feasible region is unbounded therefore, 7 may or may not be the minimum value of Z. For this, we draw the graph of the inequality, $3 x + 5 y < 7$ and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with $3 x + 5 y < 7$ . Therefore, the minimum value of Z is 7 at the point B $(\frac{3}{2}, \frac{1}{2})$ .

Maximize $Z = 3 x + 2 y$ , subject to constraints $x + 2 y \leq 10, 3 x + y \leq 15$ and $x, y \geq 0$ . Sol. We have to Maximize $Z = 3 x + 2 y$ Subject to constraints $x + 2 y \leq 10, 3 x + y \leq 15, x \geq 0, y \geq 0$ Firstly, draw the graph of the line, $x + 2 y = 10$ Putting $(0, 0)$ in the inequality $x + 2 y \leq 10$ , we have $0 + 2 \times 0 \leq 10 \Rightarrow 0 \leq 10$ (Which is true) So, the half plane is towards the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $3 x + y = 15$ Putting $(0, 0)$ in the inequality $3 x + y \leq 15$ we have $3 \times 0 + 0 \leq 15$ $\Rightarrow 0 \leq 15$ (Which is true) So, the half plane is towards the origin. On solving equations $x + 2 y = 10$ and $3 x + y = 15$ , we get $x = 4$ and $y = 3$ ∴ Intersection point $B$ is $(4, 3)$ ∴ Feasible region is OABCO.

The corner points of the feasible region are $O (0, 0), A (5, 0), B (4, 3)$ and $C (0, 5)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 2y$
$O (0, 0)$	0
A(5,0)	15
B(4,3)	18 → Maximum
C(0,5)	10

Therefore, the maximum value of Z is 18 at the point $B (4, 3)$ .

Minimize $Z = x + 2 y$ , subject to constraints are $2 x + y \geq 3, x + 2 y \geq 6$ and $x, y \geq 0$ .

Corner Points	$Z = x + 2y$
A(6,0)	6
B( 0,3)	6

Show that the minimum of Z occurs at more than two points. Sol. We have to Minimize $Z = x + 2 y$ Subject to constraints

$2 x + y \geq 3, x + 2 y \geq 6, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $2 x + y = 3$ Putting $(0, 0)$ in the inequality $2 x + y \geq 3$ , we have $2 \times 0 + 0 \geq 3 \Rightarrow 0 \geq 3$ (Which is false) So, the half plane is away from the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $x + 2 y = 6$ Putting $(0, 0)$ in the inequality $x + 2 y \geq 6$ we have $0 + 2 \times 0 \geq 6 \Rightarrow 0 \geq 6$ (Which is false) So, the half plane is away from the origin. The intersection point of the lines $x + 2 y = 6$ and $2 x + y = 3$ is $B (0, 3)$

The corner points of the feasible region are, A(6,0), B(0,3). The values of Z at these points are as follows:

As the feasible region is unbounded therefore, $Z = 6$ may or may not be the minimum value. For this, we draw the inequality, $x + 2 y < 6$ and check whether the resulting half plane has points in common with the feasible region or not. Here there is no common point between unbounded feasible region and open half plane. Therefore, the value of Z is minimum at every point on the line, $x + 2 y = 6$ .

Minimize and maximize $Z = 5 x + 10 y$ subject to constraints are

$x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0 and x, y \geq 0.$

Sol. We have to Minimize and maximize $Z = 5 x + 10 y$ Subject to constraints

$x + 2 y \leq 120, x + y \geq 60, x, y \geq 0, x - 2 y \geq 0$

Firstly, draw the graph of the line,

$x + 2 y = 120$

Putting $(0, 0)$ in the inequality $x + 2 y \leq 120$ , we have $0 + 2 \times 0 \leq 120 \Rightarrow 0 \leq 120$ (Which is true) So, the half plane is towards the origin. Secondly, draw the graph of the line, $x + y = 60$ Putting $(0, 0)$ in the inequality $x + y \geq 60$ , we have $0 + 0 \geq 60 \Rightarrow 0 \geq 60$ (Which is false) So, the half plane is away from the origin. Thirdly, draw the graph of the line $x - 2 y = 0$ Putting $(5, 0)$ in the inequality $x - 2 y \geq 0$ we have $5 - 2 \times 0 \geq 0 \Rightarrow 5 \geq 0$ (Which is true) So, the half plane is towards the $X$ -axis. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. $∴$ Feasible region is ABCDA. On solving equations

$x - 2 y = 0 and x + y = 60, we get D (40, 20)$

And on solving equations $x - 2 y = 0$ and $x + 2 y = 120$ , we getC $(60, 30)$

The corner points of the feasible region are, A(60,0), B(120,0), C(60,30) and D(40,20). The values of Z at these points are as follows:

Corner Points	$Z = 5x + 10y$
A $(60, 0)$	$300 \to$ Minimum
B $(120, 0)$	600 →Maximum
C(60, 30)	600 → Maximum
D(40, 20)	400

The minimum value of Z is 300 at A $(60, 0)$ and the maximum value of Z is 600 at all the points on the line segment joining the points B $(120, 0)$ and C $(60, 30)$ .

Minimize and maximize $Z = x + 2 y$ subject to constraints are

$x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200 and x, y \geq 0.$

Sol. We have to Minimize and maximize $Z = x + 2 y$ Subject to constraints

$x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x + 2 y = 100$ Putting $(0, 0)$ in the inequality $x + 2 y \geq 100$ , we have $0 + 2 \times 0 \geq 100 \Rightarrow 0 \geq 100$ (Which is false) So, the half plane is away from the origin. Secondly, draw the graph of the line $2 x - y = 0$ Putting $(5, 0)$ in the inequality $2 x - y \leq 0$ we have $2 \times 5 - 0 \leq 0 \Rightarrow 10 \leq 0$ (Which is false) So, the half plane is towards Y-axis. Thirdly, draw to graph of the line $2 x + y = 200$ Putting $(0, 0)$ in the inequality $2 x + y \leq 200$ we have $2 \times 0 + 0 \leq 200 \Rightarrow 0 \leq 200$ (Which is true) So, the half plane is towards the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant.

On solving equations $2 x - y = 0$ and $x + 2 y = 100$ , we get $B (20, 40)$ And on solving equations $2 x - y = 0$ and $2 x + y = 200$ , we get $C (50, 100)$

$∴$ Feasible region is ABCDA . The corner points of the feasible region are, A(0, 50), B(20, 40), C(50,100) and D(0, 200) The values of Z at these points are as follows:

Corner Points	$Z = x + 2y$
A(0, 50),	$100 \to$ Minimum
B(20, 40),	100 → Minimum
C(50,100)	250
D(0, 200)	400 →Maximum

The maximum value of Z is 400 at $D (0, 200)$ and the minimum value of Z is 100 at all the points on the line segment joining $A (0, 50)$ and B(20,40).

For this, we draw the inequality, $- x + 2 y > 1$ and check whether the resulting half plane has points in common with the feasible region or not. The resulting feasible region has points in common with the feasible region. Therefore, $Z = 1$ , is not the maximum value. Hence, Z has no maximum value.

Maximize $Z = - x + 2 y$ , subject to the constraints

$x \geq 3, x + y \geq 5, x + 2 y \geq 6 and y \geq 0.$

Sol. We have to Maximize $Z = - x + 2 y$ Subject to constraints

$x \geq 3, x + y \geq 5, x + 2 y \geq 6, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x + y = 5$ Putting $(0, 0)$ in the inequality $x + y \geq 5$ , we have $0 + 0 \geq 5 \Rightarrow 0 \geq 5$ (Which is false) So, the half plane is away from the origin. Secondly, draw the graph of the line, $x + 2 y = 6$ Putting $(0, 0)$ in the inequality $x + 2 y \geq 6$ , we have $0 + 2 \times 0 \geq 6 \Rightarrow 0 \geq 6$ (Which is false) So, the half plane is away from the origin. It can be seen that the feasible region is unbounded. The corner points of the feasible region are A(6, 0), B(4, 1) and C(3, 2). The values of Z at these points are as follows:

Corner Points	$Z = - x + 2y$
A(6, 0)	-6
B(4,1)	-2
C(3, 2)	$1 \to$ Maximum

As the feasible region is unbounded therefore, $Z = 1$ may or may not be the maximum value.

For this, we draw the inequality, $[- x + 2 y > 1]$ and check whether the resulting half plane has points in common with the feasible region or not. The resulting feasible region has points in common with the feasible region. Therefore, Z = 1, is not the maximum value.

Hence, Z has no maximum value.

Maximize $Z = x + y$ , subject to constraints are

$x - y \leq - 1, - x + y \leq 0 and x, y \geq 0.$

Sol. We have to Maximize $Z = x + y$ Subject to constraints $x - y \leq - 1, - x + y \leq 0$ ,

$x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x - y = - 1$ Putting $(0, 0)$ in the inequality $x - y \leq - 1$ , we have $0 - 0 \leq - 1 \Rightarrow 0 \leq - 1$ (Which is false) So, the half plane is away from the origin. Secondly, draw the graph of the line,

$- x + y = 0$

Putting $(2, 0)$ in the inequality $- x + y \leq 0$ we have $- 2 + 0 \leq 0 \Rightarrow - 2 \leq 0$ (Which is true) So, the half plane is towards the $X$ -axis. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant.

From the above graph, it is clearly shown that there is no common region. Hence, there is no feasible region and thus Z has no maximum value.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Maths chapter-wise, covering exercise questions, useful formulas, and step-by-step solutions to help students understand concepts and practise effectively.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 13	Probability

4.0Class 12 Maths Chapter 12 Linear Programming Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 12.1	10 Questions and Solutions	Formulation of linear programming problems and graphical method of solving them

5.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Class 12 Maths Chapter 12, Linear Programming, focuses on solving optimisation problems using graphical methods. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 12 include:

Linear Programming Problem (LPP): Learn how real-life situations can be expressed as mathematical optimisation problems.
Objective Function: Understand how to define a function that needs to be maximised or minimised based on the given problem.
Constraints: Study the linear inequalities that represent restrictions and limit the possible solutions.
Feasible Region: Identify the common region on a graph that satisfies all the given constraints.
Corner Point Method: Find the optimal solution by evaluating the objective function at the corner points of the feasible region.
Graphical Representation: Learn how to draw and interpret graphs of linear inequalities to identify the feasible region and solve LPP questions.

NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

1.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Class 12 Maths Chapter 12, Linear Programming, focuses on solving optimisation problems using graphical methods. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 12 include:

Linear Programming Problem (LPP): Learn how real-life situations can be expressed as mathematical optimisation problems.
Objective Function: Understand how to define a function that needs to be maximised or minimised based on the given problem.
Constraints: Study the linear inequalities that represent restrictions and limit the possible solutions.
Feasible Region: Identify the common region on a graph that satisfies all the given constraints.
Corner Point Method: Find the optimal solution by evaluating the objective function at the corner points of the feasible region.
Graphical Representation: Learn how to draw and interpret graphs of linear inequalities to identify the feasible region and solve LPP questions.

2.0NCERT Class 12 Maths Chapter 12 Linear Programming : Detailed Solutions

EXERCISE - 12.1

Solve the following Linear Programming problems graphically.

Maximize $Z = 3 x + 4 y$ , Subject to the constraints

$x + y \leq 4, x \geq 0 and y \geq 0.$

∴ Feasible region is OABO. The corner points of the feasible region are $O (0, 0), A (4, 0)$ and $B (0, 4)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 4y$
O(0,0)	0
A(4,0)	12
B(0,4)	$16 \to$ Maximum

Therefore, the maximum value of Z is 16 at the point B(0,4).

Minimize $Z = - 3 x + 4 y$ , subject to constraints

$x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0 and y \geq 0 .$

Sol. We have to Minimize $Z = - 3 x + 4 y$ Subject to constraints

$x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$

Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $3 x + 2 y = 12$

x	0	4
y	6	0

$x + 2 y = 8 and 3 x + 2 y = 12,$

we get $x = 2$ and $y = 3$ ∴ Intersection point B is $(2, 3)$ The corner points of the feasible region are $O (0, 0), A (4, 0), B (2, 3)$ and $C (0, 4)$ .

The values of Z at these points are as follows.

Corner Points	$Z = - 3x + 4y$
$O (0, 0)$	0
A(4,0)	-12 →Minimum
B(2,3)	6
C(0,4)	16

Therefore, the minimum value of Z is -12 at the point A(4,0).

Maximize $Z = 5 x + 3 y$ , subject to constraints

$3 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0 and y \geq 0.$

Sol. We have to Maximize $Z = 5 x + 3 y$ Subject to constraints

$3 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0, y \geq 0$

Firstly, draw the graph of the line,

$3 x + 5 y = 15$

Putting $(0, 0)$ in the inequality $3 x + 5 y \leq 15$ , we have

$3 \times 0 + 5 \times 0 \leq 15$

$3 x + 5 y = 15 and 5 x + 2 y = 10$

we get $x = \frac{20}{19}$ and $y = \frac{45}{19}$

Coordinates of point B is $(\frac{20}{19}, \frac{45}{19})$

∴ Feasible region is OABCO The corner points of the feasible region are $O (0, 0), A (2, 0), B (\frac{20}{19}, \frac{45}{19})$ and $C (0, 3)$ . The values of Z at these points are as follows:

Corner Points	$Z = 5x + 3y$
$O (0, 0)$	0
A(2,0)	10
$B (\frac{20}{19}, \frac{45}{19})$	$\frac{235}{19} \to$ Maximum
C(0,3)	9

Therefore, the maximum value of Z is $\frac{235}{19}$ at the point $B (\frac{20}{19}, \frac{45}{19})$ .

Minimize $Z = 3 x + 5 y$ subject to constraints

$x + 3 y \geq 3, x + y \geq 2 and x, y \geq 0.$

Sol. We have to Minimize $Z = 3 x + 5 y$ Subject to constraints

$x + 3 y \geq 3, x + y \geq 2, x \geq 0, y \geq 0$

Firstly, draw the graph of the line, $x + 3 y = 3$ Putting $(0, 0)$ in the inequality $x + 3 y \geq 3$ , we have $0 + 3 \times 0 \geq 3 \Rightarrow 0 \geq 3$ (Which is false)

The corner points of the feasible region are $A (3, 0), B (\frac{3}{2}, \frac{1}{2})$ and $C (0, 2)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 5y$
A(3,0)	9
B $(\frac{3}{2}, \frac{1}{2})$	$7 \to$ Minimum
C(0,2)	10

It can be seen that the feasible region has no common point with $3 x + 5 y < 7$ . Therefore, the minimum value of Z is 7 at the point B $(\frac{3}{2}, \frac{1}{2})$ .

Maximize $Z = 3 x + 2 y$ , subject to constraints $x + 2 y \leq 10, 3 x + y \leq 15$ and $x, y \geq 0$ . Sol. We have to Maximize $Z = 3 x + 2 y$ Subject to constraints $x + 2 y \leq 10, 3 x + y \leq 15, x \geq 0, y \geq 0$ Firstly, draw the graph of the line, $x + 2 y = 10$ Putting $(0, 0)$ in the inequality $x + 2 y \leq 10$ , we have $0 + 2 \times 0 \leq 10 \Rightarrow 0 \leq 10$ (Which is true) So, the half plane is towards the origin. Since, $x, y \geq 0$ So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line, $3 x + y = 15$ Putting $(0, 0)$ in the inequality $3 x + y \leq 15$ we have $3 \times 0 + 0 \leq 15$ $\Rightarrow 0 \leq 15$ (Which is true) So, the half plane is towards the origin. On solving equations $x + 2 y = 10$ and $3 x + y = 15$ , we get $x = 4$ and $y = 3$ ∴ Intersection point $B$ is $(4, 3)$ ∴ Feasible region is OABCO.

The corner points of the feasible region are $O (0, 0), A (5, 0), B (4, 3)$ and $C (0, 5)$ . The values of Z at these points are as follows:

Corner Points	$Z = 3x + 2y$
$O (0, 0)$	0
A(5,0)	15
B(4,3)	18 → Maximum
C(0,5)	10

Therefore, the maximum value of Z is 18 at the point $B (4, 3)$ .

Minimize $Z = x + 2 y$ , subject to constraints are $2 x + y \geq 3, x + 2 y \geq 6$ and $x, y \geq 0$ .

Corner Points	$Z = x + 2y$
A(6,0)	6
B( 0,3)	6

Show that the minimum of Z occurs at more than two points. Sol. We have to Minimize $Z = x + 2 y$ Subject to constraints

$2 x + y \geq 3, x + 2 y \geq 6, x \geq 0, y \geq 0$

The corner points of the feasible region are, A(6,0), B(0,3). The values of Z at these points are as follows:

Minimize and maximize $Z = 5 x + 10 y$ subject to constraints are

$x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0 and x, y \geq 0.$

Sol. We have to Minimize and maximize $Z = 5 x + 10 y$ Subject to constraints

$x + 2 y \leq 120, x + y \geq 60, x, y \geq 0, x - 2 y \geq 0$

Firstly, draw the graph of the line,

$x + 2 y = 120$

$x - 2 y = 0 and x + y = 60, we get D (40, 20)$

And on solving equations $x - 2 y = 0$ and $x + 2 y = 120$ , we getC $(60, 30)$

The corner points of the feasible region are, A(60,0), B(120,0), C(60,30) and D(40,20). The values of Z at these points are as follows:

Corner Points	$Z = 5x + 10y$
A $(60, 0)$	$300 \to$ Minimum
B $(120, 0)$	600 →Maximum
C(60, 30)	600 → Maximum
D(40, 20)	400

The minimum value of Z is 300 at A $(60, 0)$ and the maximum value of Z is 600 at all the points on the line segment joining the points B $(120, 0)$ and C $(60, 30)$ .

Minimize and maximize $Z = x + 2 y$ subject to constraints are

$x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200 and x, y \geq 0.$

Sol. We have to Minimize and maximize $Z = x + 2 y$ Subject to constraints

$x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x \geq 0, y \geq 0$

On solving equations $2 x - y = 0$ and $x + 2 y = 100$ , we get $B (20, 40)$ And on solving equations $2 x - y = 0$ and $2 x + y = 200$ , we get $C (50, 100)$

$∴$ Feasible region is ABCDA . The corner points of the feasible region are, A(0, 50), B(20, 40), C(50,100) and D(0, 200) The values of Z at these points are as follows:

Corner Points	$Z = x + 2y$
A(0, 50),	$100 \to$ Minimum
B(20, 40),	100 → Minimum
C(50,100)	250
D(0, 200)	400 →Maximum

The maximum value of Z is 400 at $D (0, 200)$ and the minimum value of Z is 100 at all the points on the line segment joining $A (0, 50)$ and B(20,40).

Maximize $Z = - x + 2 y$ , subject to the constraints

$x \geq 3, x + y \geq 5, x + 2 y \geq 6 and y \geq 0.$

Sol. We have to Maximize $Z = - x + 2 y$ Subject to constraints

$x \geq 3, x + y \geq 5, x + 2 y \geq 6, x \geq 0, y \geq 0$

Corner Points	$Z = - x + 2y$
A(6, 0)	-6
B(4,1)	-2
C(3, 2)	$1 \to$ Maximum

As the feasible region is unbounded therefore, $Z = 1$ may or may not be the maximum value.

Hence, Z has no maximum value.

Maximize $Z = x + y$ , subject to constraints are

$x - y \leq - 1, - x + y \leq 0 and x, y \geq 0.$

Sol. We have to Maximize $Z = x + y$ Subject to constraints $x - y \leq - 1, - x + y \leq 0$ ,

$x \geq 0, y \geq 0$

$- x + y = 0$

From the above graph, it is clearly shown that there is no common region. Hence, there is no feasible region and thus Z has no maximum value.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Maths chapter-wise, covering exercise questions, useful formulas, and step-by-step solutions to help students understand concepts and practise effectively.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 13	Probability

4.0Class 12 Maths Chapter 12 Linear Programming Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 12.1	10 Questions and Solutions	Formulation of linear programming problems and graphical method of solving them

5.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Class 12 Maths Chapter 12, Linear Programming, focuses on solving optimisation problems using graphical methods. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 12 include:

Linear Programming Problem (LPP): Learn how real-life situations can be expressed as mathematical optimisation problems.
Objective Function: Understand how to define a function that needs to be maximised or minimised based on the given problem.
Constraints: Study the linear inequalities that represent restrictions and limit the possible solutions.
Feasible Region: Identify the common region on a graph that satisfies all the given constraints.
Corner Point Method: Find the optimal solution by evaluating the objective function at the corner points of the feasible region.
Graphical Representation: Learn how to draw and interpret graphs of linear inequalities to identify the feasible region and solve LPP questions.

Frequently Asked Questions

How do NCERT Solutions for Class 12 Maths Chapter 12 explain the formulation of an LPP?

What is a feasible region according to NCERT Class 12 Maths Chapter 12 Solutions?

Why is the corner point method used in NCERT Solutions for Class 12 Maths Chapter 12?

What should I practise from Exercise 12.1 using NCERT Solutions for Class 12 Maths Chapter 12?

How can NCERT Solutions for Class 12 Maths Chapter 12 help with graph-based LPP questions?

What is the objective function in NCERT Class 12 Maths Chapter 12 Linear Programming Solutions?

Can NCERT Solutions for Class 12 Maths Chapter 12 help with competitive exams like JEE?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

1.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

2.0NCERT Class 12 Maths Chapter 12 Linear Programming : Detailed Solutions

EXERCISE - 12.1

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 12 Linear Programming Exercise-wise Solutions

5.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Table of Contents

Frequently Asked Questions

How do NCERT Solutions for Class 12 Maths Chapter 12 explain the formulation of an LPP?

What is a feasible region according to NCERT Class 12 Maths Chapter 12 Solutions?

Why is the corner point method used in NCERT Solutions for Class 12 Maths Chapter 12?

What should I practise from Exercise 12.1 using NCERT Solutions for Class 12 Maths Chapter 12?

How can NCERT Solutions for Class 12 Maths Chapter 12 help with graph-based LPP questions?

What is the objective function in NCERT Class 12 Maths Chapter 12 Linear Programming Solutions?

Can NCERT Solutions for Class 12 Maths Chapter 12 help with competitive exams like JEE?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

1.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

2.0NCERT Class 12 Maths Chapter 12 Linear Programming : Detailed Solutions

EXERCISE - 12.1

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 12 Linear Programming Exercise-wise Solutions

5.0Key Concepts of Class 12 Maths Chapter 12 Linear Programming

Table of Contents