Question

Solve the following linear programming problem graphically.

Maximize Z = 60x₁ + 15x₂ subject to the constraints: x₁ + x₂ ≤ 50; 3x₁ + x₂ ≤ 90 and x₁, x₂ ≥ 0.

Answer 1

Since the decision variables, x₁ and x₂ are non-negative, the solution lies in the I quadrant of the plane.

Consider the equations

x₁ + x₂ = 50

x1	0	50
x2	50	0

3x₁ + x₂ = 90

x1	0	30
x2	90	0

The feasible region is OABC and its co-ordinates are O(0, 0) A(30, 0) C(0, 50) and B is the point of intersection of the lines

x₁ + x₂ = 50 ..........(1)

3x₁ + x₂ = 90 .........(2)

Verification of B:

x₁ + x₂ = 50 ..........(1)
3x₁ + x₂ = 90 .........(2)
−     −       −       
− 2x₁ = − 40

x₁ = 20

From (1), 20 + x₂ = 50

x₂ = 30

∴ B is (20, 30)

Corner points	Z = 60x1 + 15x2
O(0, 0)	0
A(30, 0)	1800
B(20, 30)	1650
C(0, 50)	7500

Maximum value occurs at C(0, 50)

∴ The solution is x₁ = 0, x₂ = 50 and Z_max = 7500.