Maximize Z = 40x1 + 50x2 subject to constraints 3x1 + x2 ≤ 9; x1 + 2x2 ≤ 8 and x1, x2 ≥ 0.

Question

Solve the linear programming problem by graphical method.

Maximize Z = 40x₁ + 50x₂ subject to constraints 3x₁ + x₂ ≤ 9; x₁ + 2x₂ ≤ 8 and x₁, x₂ ≥ 0.

Answer 1

Given that 3x₁ + x₂ ≤ 9

Let 3x₁ + x₂ = 9

x1	0	3
x2	9	0

Also given that x₁ + 2x₂ ≤ 8

Let x₁ + 2x₂ = 8

x1	0	8
x2	4	0

3x₁ + x₂ = 9 … (1)

x₁ + 2x₂ = 8 … (2)

(1) x 2 ⇒ 6x₁ + 2x₂ = 18 ... (3)

(2) + (3) ⇒ -5x₁ = -10

x₁ = 2

x₁ = 2 substitute in (1)

3(2) + x₂ = 9

x₂ = 3

The feasible region satisfying all the conditions is OABC.

The co-ordinates of the corner points are O(0, 0), A(3, 0), B(2, 3), C(0, 4)

The maximum value of Z occurs at (2, 3).

∴ The optimal solution is x₁ = 2, x₂ = 3 and Z_max = 230