Question

Solve the following linear programming problems 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
x2	25	0

3x₁ + 3x₂ = 36

x₁ + x₂ = 12 ……….(1)

5x₁ + 2x₂ = 50 ………(2)

2x₁ + 2x₂ = 24 ....[(1) × 2]

−   −        −     
−3x₁ = − 6

x₁ = 2

Substituting x₁ = 2 in (1) we get

2+ x₂ = 12

x₂ = 6

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)
– 2x₂ = – 18 ......[Equation (1) – (2)]

x₂ = 9

x₂ = 9 substitute in (1)

x₁ + x₂ = 12

x₁ + 9 = 12

x₁ = 12 – 9

x₁ = 3

The feasible region satisfying all the given conditions is OABCD.

The co-ordinates of the comer points are

Corner points	Z = 20x1 + 30x2
O(0, 0)	0
A(10, 0)	200
B(2, 6)	220
C(3, 9)	330
D(0, 10)	300

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

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