Find out the corner point P in the feasible region of the LPP in which the objective function z = - 2 x + 6 y is maximum subject to the constraints 3 x + 5 y ≥ 15 , x − y + 3 ≥ 0 , x − 3 y + 15 ≥ 0 , where x ≥ 0 , y ≥ 0 . Determine also the maximum value of z . Does there exist the minimum value of z with respect to same set of constraints? Answer with proper reason.

Given LPP is as follows: Objective function: z = - 2 x + 6 y Subject to 3 x + 5 y ≥ 15 x - y + 3 ≥ 0 x - 3 y + 15 ≥ 0 , where x ≥ 0 , y ≥ 0 Let us first find the feasible region of the LPP We first draw the graph of the line 3 x + 5 y ≥ 15 on the x y - plane by joining the points 0 , 3 , and 5 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 3 x + 5 y ≥ 15 . Since the coordinates of the point 0 , 0 does not satisfy the inequation 3 x + 5 y ≥ 15 . So, the solution region for this inequation is that half-plane which does not contain the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Similarly, we can find the solution region for x - y + 3 ≥ 0 and x - 3 y + 15 ≥ 0 Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region of the problem is unbounded and has three corner points which is at A 0 , 3 , B 5 , 0 , C 3 , 6 . The values of Z at these corner points are as follows: Corner Point ( x , y ) Value of z at ( x , y ) A ( 0 , 3 ) z = 0 + 18 = 18 B ( 5 , 0 ) z = - 10 + 0 = - 10 C ( 3 , 6 ) z = - 6 + 36 = 30 → maximum As the feasible region is unbounded, therefore, - 10 may be or may not be the minimum value of z . For this, we draw the graph of the inequality, - 2 x + 6 y < - 10 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 = - 10 is not the minimum value. z has no minimum value. As the feasible region is unbounded, therefore, 30 may be or may not be the maximum value of z . For this, we draw the graph of the inequality, - 2 x + 6 y > 30 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 - 2 x + 6 y > 30 Therefore, the maximum value of z is 30 at 3 , 6 .

The constraints of an LPP with two decision variables x , y are given to be y ≤ 3 x , 3 x + 4 y ≥ 15 and 2 x + y ≤ 10 where x ≥ 0 , y ≥ 0 . The objective functions z of this LPP is maximum at the point 3 . 5 , 3 . Determine z if max z = 30 .

Given that an LPP is subject to the constraints y ≤ 3 x , 3 x + 4 y ≥ 15 and 2 x + y ≤ 10 , where x ≥ 0 , y ≥ 0 Let the objective function of an LPP be z = a x + b y Then, maximum value of z is 30 . Let us first find the feasible region of the given constraints. The common shaded region in the first quadrant represents the solution set of the given constraints Since the maximum value of an objective function is attained at the point ( 3 . 5 , 3 ) that lies on the line segment joining two corner points 5 , 0 , and 2 , 6 i.e., mid-point of the line segment B C We know that if two corner points of the feasible region are optimal solutions of the same type, i.e., both produce the same maximum or minimum, then any point on the line segment joining these two points is also an optimal solution of the same type. Thus, the optimal solution of the given LPP occurs at infinitely many points that lies on the line segment joining the corner points ( 5 , 0 ) , and ( 2 , 6 ) . Now, the coordinates of these points will satisfy the objective function, z = a x + b y = 30 ( maximum value ) So, satisfying the coordinates of the point ( 3 . 5 , 3 ) , we have: 30 = 7 2 a + 3 b ⇒ 7 a + 6 b = 60 → 1 Now, satisfying the coordinates of point 2 , 6 , we have: 30 = 2 a + 6 b → 2 Applying, 1 - 2 , we get: 5 a = 30 ⇒ a = 6 Thus, 30 = 12 + 6 b ⇒ b = 3 (using 2 ) Hence, z = a x + b y ⇒ z = 6 x + 3 y Hence, the required objective function is z = 6 x + 3 y .

Maximize z = x + y subject to x + 2 y ≤ 10 x + y ≥ 1 y ≤ 4 where, x , y ≥ 0 .

Given LPP as follows: Objective function: z = x + y Subject to x + 2 y ≤ 10 x + y ≥ 1 y ≤ 4 , where x ≥ 0 , y ≥ 0 Let us first find the feasible region of the LPP We first draw the graph of the line x + 2 y ≤ 10 on the x y - plane by joining the points 0 , 5 and 10 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation x + 2 y ≤ 10 . Since the coordinates of the point 0 , 0 satisfy the inequation x + 2 y ≤ 10 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Similarly, we can find the solution region for x + y ≥ 1 and y ≤ 4 . Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. Thus, the feasible region of the given system of constraints is represented by the shaded region in the following graph: The corner points of the feasible region are A 1 , 0 , B 10 , 0 , C 4 , 2 , and D - 3 , 1 . The values of Z at these corner points are as follows: Corner Point ( x , y ) Value of z at ( x , y ) A ( 1 , 0 ) z = 1 + 0 = 1 B ( 10 , 0 ) z = 10 + 0 = 10 → maximum C ( 4 , 2 ) z = 4 + 2 = 6 D ( - 3 , 1 ) z = - 3 + 1 = - 2 Since the feasible region is bounded. Hence, the maximum value of z is 10 that occurs at the point 10 , 0 .

Minimize z = 3 x + 2 y subject to 2 x + y ≥ 14 2 x + 3 y ≥ 22 x + y ≥ 5 where x , y ≥ 0 .

Given LPP as follows: Objective function: z = 3 x + 2 y Subject to 2 x + y ≥ 14 2 x + 3 y ≥ 22 x + y ≥ 5 , where x ≥ 0 , y ≥ 0 Let us first find the feasible region of the LPP We first draw the graph of the line 2 x + y ≥ 14 on the x y - plane by joining the points 0 , 14 and 7 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 2 x + y ≥ 14 . Since the coordinates of the point 0 , 0 does satisfy the inequation 2 x + y ≥ 14 . So, the solution region for this inequation is that half-plane which does not contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Similarly, we can find the solution region for 2 x + 3 y ≥ 22 and x + y ≥ 5 . Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. The feasible region of the problem is unbounded and has three corner points which is at A 0 , 14 , B 5 , 4 , C 11 , 0 . The values of Z at these corner points are as follows: Corner Point ( x , y ) Value of z at ( x , y ) A ( 0 , 14 ) z = 0 + 28 = 28 B ( 5 , 4 ) z = 15 + 8 = 23 → minimum C ( 11 , 0 ) z = 33 + 0 = 33 As the feasible region is unbounded, therefore, 23 may be or may not be the minimum value of z . For this, we draw the graph of the inequality, 3 x + 2 y < 23 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 + 2 y < 23 Hence, the minimum value of z is 23 that occurs at the point 5 , 4 .

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 2

Using iso-profit (or iso-cost) method, determine the optimal solutions of the following LPP's. Find also the optimum value of the objective function in each case.

Maximise $z = - x - 2 y$

Subject to $7 x + 8 y \leq 56$ ,

$2 x + y \geq 6$ ,

where $x \geq 0, y \geq 0$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 3

Determine all the points in the feasible region at which the maximum values of the objective function $z = 12 x + 2 y$ is attained subject to the constraints $B (8, 0) : Z = - 8 - 2 (0) = - 8$ where $x \geq 0, y \geq 0$ . Find also the maximum value of $z$

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 4

Find all the optimal solution of the LPP in which the objective function $z = - 16 x - 4 y$ is to be minimized subject to the constraints $4 x + y \leq 36, 2 x - 7 y + 42 \leq 0, 4 x + 3 y \geq 12$ where $x \geq 0, y \geq 0$ . Determine also the minimum value of $z$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 5

Determine the maximum and minimum values (if exists) of the objective function $z = 9 x - 4 y$ subject to the constraints $3 x - 2 y \geq 6, 3 x - y \geq 9, x - 7 y \leq 7$ where $x \geq 0, y \geq 0$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 6

Find out the corner point $P$ in the feasible region of the LPP in which the objective function $z = - 2 x + 6 y$ is maximum subject to the constraints $3 x + 5 y \geq 15, x - y + 3 \geq 0, x - 3 y + 15 \geq 0$ , where $x \geq 0, y \geq 0$ . Determine also the maximum value of $z$ . Does there exist the minimum value of $z$ with respect to same set of constraints? Answer with proper reason.

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 7

The constraints of an LPP with two decision variables $x, y$ are given to be $y \leq 3 x, 3 x + 4 y \geq 15$ and $2 x + y \leq 10$ where $x \geq 0, y \geq 0$ . The objective functions $z$ of this LPP is maximum at the point $(3.5, 3)$ . Determine $z$ if max $z = 30$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 9

Maximize $z = x + y$

subject to $x + 2 y \leq 10$

$x + y \geq 1$

$y \leq 4$

where, $x, y \geq 0$ .

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 26 - Linear Programming Problems>EXERCISE>Q 10

Minimize $z = 3 x + 2 y$

subject to $2 x + y \geq 14$

$2 x + 3 y \geq 22$

$x + y \geq 5$

where $x, y \geq 0$ .

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

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