Maximize Z = 22x1 + 18x2 subject to constraints 960x1 + 640x2 ≤ 15360; x1 + x2 ≤ 20 and x1, x2 ≥ 0.

Question

Solve the linear programming problem by graphical method.

Maximize Z = 22x₁ + 18x₂ subject to constraints 960x₁ + 640x₂ ≤ 15360; x₁ + x₂ ≤ 20 and x₁, x₂ ≥ 0.

Answer 1

Given that 960x₁ + 640x₂ ≤ 15360

Let 960x₁ + 640x₂ = 15360

3x₁ + 2x₂ = 48

x1	0	16
x2	24	0

Also given that x₁ + x₂ ≤ 20

Let x₁ + x₂ = 20

x1	0	20
x2	20	0

To get point of intersection

3x₁ + 2x₂= 48 … (1)

x₁ + x₂ = 20 ... (2)

(2) x -2 ⇒ -2x₁ – 2x₂ = -40 … (3)

(1) + (3) ⇒ x₁ = 8

x₁ = 8 substitute in (2),

8 + x₂ = 20

x₂ = 12

The feasible region satisfying all the given conditions is OABC.

The co-ordinates of the comer points are O(0, 0), A(16, 0), B(8,12) and C(0, 16).

The maximum value of Z occurs at B(8, 12).

∴ The optimal solution is x₁ = 8, x₂ = 12 and Z_max = 392