Question

Solve the following L.P.P. by graphical method:

Min Z = 20x₁ + 10x₂

Subject to x₁ + 2x₂ ≤ 40

3x₁ + x₂ ≥ 30

4x₁ + 3x₂ ≥ 60

x₁, x₂ ≥ 0

Answer 1

Convert all the inequalities of the constraints into equations, we have

x₁ + 2x₂ = 40

3x₁ + x₂ = 30

4x₁ + 3x₂ = 60

x₁ + 2x₂ = 40 passes through (0, 20) (40, 0)

3x₁ + x₂ = 30 passes through (0, 30) (10, 0)

4x₁ + 3x₂ = 60 passes through (0, 20) (15, 0)

Plot above equations on graph, we have

Here feasible region is ABCD.

The coordinates of ABCD are A(15, 0) B(40, 0), C(4, 18), D(6, 12)

Now

Corner Points	Coordinate	Coordinate
A	(15, 0)	300
B	(40, 0)	800
C	(4, 18)	260
D	(6, 12)	240

Therefore, minimum value of Z occurs at D(6, 12).

Hence, optimal solution is x₁ = 6, x₂ = 12.