Maximize Z = 20x1 + 30x2 subject to constraints 3x1 + 3x2 ≤ 36; 5x1 + 2x2 ≤ 50; 2x1 + 6x2 ≤ 60 and x1, x2 ≥ 0.

Question

Solve the linear programming problem by graphical method.

Maximize Z = 20x₁ + 30x₂ subject to constraints 3x₁ + 3x₂ ≤ 36; 5x₁ + 2x₂ ≤ 50; 2x₁ + 6x₂ ≤ 60 and x₁, x₂ ≥ 0.

Answer 1

Given that 3x₁ + 3x₂ ≤ 36

Let 3x₁ + 3x₂ = 36

x1	0	12
x2	12	0

Also given that 5x₁ + 2x₂ ≤ 50

Let 5x₁ + 2x₂ = 50

x1	0	10	6
x2	25	0	10

x₁ + x₂ = 12 ... (1)

5x₁ + 2x₂ = 50 … (2)

(1) x 2 ⇒ 2x₁ + 2x₂ = 24 … (3)

(2) – (3) ⇒ 3x₁ = 26

x₁ = \(\frac{26}{3}\) = 8.66

put x₁ = \(\frac{26}{3}\) substitute in (1)

x₁ + x₂ = 12

x₂ = 12 – x₁

x₂ = 26 – \(\frac{26}{3}\) = \(\frac{10}{3}\) = 3.33

Also given that 2x₁ + 6x₂ ≤ 60

Let 2x₁ + 6x₂ = 60

x₁ + 3x₂ = 30

x1	0	30
x2	10	0

x₁ + x₂ = 12 … (1)

x₁ + 3x₂ = 30 … (2)

(1) – (2) ⇒ -2x₂ = -18

x₂ = 9

x₂ = 9 substitute in (1) ⇒ x₁ = 3

The feasible region satisfying all the given conditions is OABCD.

The co-ordinates of the comer points are O(0, 0), A(10, 0), B(\(\frac{26}{3},\frac{10}{3}\)), and C = (3, 9) and D (0, 10)

The maximum value of Z occurs at C(3, 9)

∴ The optimal solution is x₁ = 3, x₂ = 9 and Z_max = 330