Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Ex 6.2 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.2

Solve the following LPP by graphical method.

Question 1.

Maximize z = 11x + 8y, Subject to x ≤ 4 ,y ≤ 6 x + y ≤ 6, x ≥ 0, y ≥ 0.

Solution:

The feasible solution is AOBE

Where A(4, 0) O(0, 0) B(0, 6)

E is the point of intersection of x + y = 6 and x = 4.

∴ 4 + y = 6

∴ y = 2

∴ E = (4, 2)

∴ z is maximum at (4, 12) and the maximum value of z = 60

Question 2.

Maximize z = 4x + 6y, Subject to 3x + 2y ≤ 12, x + y ≥ 4 x, y ≥ 0.

Solution:

From figure, ABC is the feasible region

Where A(0, 6) B(4, 0) C(0, 4)

Maximum value of z = 36 at A(0, 6)

Question 3.

Maximize z = 7x + 11y, Subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.

Solution:

∴ AODE is the feasible region where

A(0, 5.2) O(0, 0) D(6, 0) and E is the intersection of 3x + 5y = 26 and 5x + 3y = 30

For E,

Solving 3x + 5y = 26 ……(i)

5x + 3y = 30 ……(ii)

We get, x = 4.5, y = 2.5

∴ E = (4.5, 2.5)

∴ Maximum value of z = 59 at E(4.5, 2.5)

Question 4.

Maximize z = 10x + 25y, Subject to 0 ≤ x ≤ 3, 0≤ y ≤ 3, x + y ≤ 5.

Solution:

The constraints can be written as, x ≤ 3, x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 5

ABCDE is the feasible region where A(3, 0) B(0, 0) and C(0, 3) D is the intersection of y = 3 and x + 5y = 5 and E is the intersection of x = 3 and x + 7 = 5

For D,

Solving y = 3 ………(i)

x + y = 5 ……..(ii)

We get x = 2, y = 3

∴ D = (2, 3)

For E,

Solving x = 3 …….(i)

x + y = 5 ……….(ii)

We get x = 3, y = 2

∴ E = (3, 2)

∴ Maximum value of z = 90 at D(2, 3)

Question 5.

Maximize z = 3x + 5y, Subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.

Solution:

OAGHD is the feasible region where O(0, 0), A(0, 6), D(7, 0) G is the intersecting point of x + 4y = 24 and x + y = 9

H is the intersecting points of 3x + y = 21 and x + y = 9.

For G, Solving x + 4y = 24 …….(i)

x + y = 9 ………(ii)

We get, x = 4, y = 5

∴ G (4, 5)

For H, Solving x + y = 9 ………(i)

3x + y = 21 ……..(ii)

We get x = 6, y = 3

∴ H(6, 3)

∴ Maximum value of z = 37 at the point G(4, 5)

Question 6.

Minimize z = 7x + y Subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0

Solution:

AED is the feasible region where A(0, 5) D(3, 0) and E is the point of intersection of 5x + y = 5 and x + y = 3.

For E, Solving 5x + y = 5 ………(i)

x + y = 3 ……..(ii)

We get, x = \(\frac{1}{2}\), y = \(\frac{5}{2}\)

∴ E(\(\frac{1}{2}\), \(\frac{5}{2}\))

∴ Minimum value of z = 5 at A(0, 5)

Question 7.

Minimize z = 8x + 10y, Subject to 2x + y ≥ 7, 2x + 3y ≥ 15 ,y ≥ 2, x ≥ 0, y ≥ 0

Solution:

AEG is the feasible solution where A(0, 7)

F is the point of intersection of 2x + y = 7 and 2x + 3y = 15

G is the point of intersection of y = 2 and 2x + 3y = 15

For F, Solving 2x + y = 7 ……..(i)

2x + 3y = 15 ……..(ii)

We get x = \(\frac{3}{2}\), y = 4

∴ F = (\(\frac{3}{2}\), 4)

For G, Solving 2x + 3y = 15 ……..(i)

y = 2 …….(ii)

We get x = 4.5, y = 2

∴ G = (4.5, 2)

∴ Minimum value of z = 52 at F(\(\frac{3}{2}\), 4)

Question 8.

Minimize z = 6x + 2y, Subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.

Solution:

DGHB is the feasible region where D(0, 3), B(4, 0)

G is the point of intersection of 3x + y = 3 and x + 2y = 3 and H is the point of intersection of x + 2y = 3 and x + 4y = 4

For G, Solving 3x + y = 3 ……(i)

x + 2y = 3 ………(ii)

W e get x = \(\frac{3}{5}\), y = \(\frac{6}{5}\)

∴ G(\(\frac{3}{5}\), \(\frac{6}{5}\))

For H, Solving x + 2y = 3 ……..(i)

x + 4y = 4 ………(ii)

We get, x = 2, y = \(\frac{1}{2}\)

∴ H(2, \(\frac{1}{2}\))

∴ Minimum value of z = 22.5 at H(2, \(\frac{1}{2}\))