Maximize Z = 6x1 + 8x2 subject to constraints 30x1 + 20x2 ≤ 300; 5x1 + 10x2 ≤ 110; and x1, x2 ≥ 0.

Question

Solve the linear programming problem by graphical method.

Maximize Z = 6x₁ + 8x₂ subject to constraints 30x₁ + 20x₂ ≤ 300; 5x₁ + 10x₂ ≤ 110; and x₁, x₂ ≥ 0.

Answer 1

Given that 30x₁ + 20x₂ ≤ 300

Let 30x₁ + 20x₂ = 300

Therefore

3x₁ + 2x₂ = 30

x1	0	10
x2	15	0

Also given that 5x₁ + 10x₂ ≤ 110

Let 5x₁ + 10x₂ = 110

x₁ + 2x₂ = 22

x1	0	22
x2	11	0

To get point of intersection, (i.e., the to get eo-ordinates of B)

3x₁ + 2x₂ = 30 … (1)

x₁ + 2x₂ = 22 … (2)

(1) – (2) ⇒ 2x₁ = 8

x₁ = 4

x₁ = 4 substitute in (1),

x₁ + 2x₂ = 22

4 + 2x₂ = 22

2x₂ = 18

x₂ = 9

i.e., B is (4, 9)

The feasible region satisfying all the given conditions is OABC.

The co-ordinates of the points are O(0, 0), A(10, 0), B(4, 9), C(0, 11).

The maximum value of Z occurs at B.

∴ The optimal solution is x₁ = 4, x₂ = 9 and Z_max = 96.