Question

Solve the following linear programming problems by graphical method.

Minimize Z = 3x₁ + 2x₂ subject to the constraints 5x₁ + x₂ ≥ 10; x₁ + x₂ ≥ 6; x₁ + 4x₂ ≥ 12 and x₁, x₂ ≥ 0.

Answer 1

Given that 5x₁ + x₂ ≥ 10

Let 5x₁ + x₂ = 10

x1	0	2
x2	10	0

Also given that x₁ + x₂ ≥ 6

Let x₁ + x₂ = 6

x1	0	6
x2	6	0

Also given that x₁ + 4x₂ ≥ 12

Let x₁ + 4x₂ = 12

x1	0	12
x2	3	0

To get B

5x₁ + x₂ = 10 ……..(1)

x₁ + x₂ = 6 ………(2)

4x₁ = 4 ......[Equation (1) – (2)]

x₁ = 1

x = 1 substitute in (2)

x₁ + x₂ = 6

1 + x₂ = 6

x₂ = 5

∴ B is (1, 5)

To get C

x₁ + x₂ = 6

x₁ + 4x₂ = 12
− 3x₂ = − 6 ..........[Equation (1) – (2)]

x₂ = 2

x₂ = 2 substitute in (2) we get,

x₁ + x₂ = 6

x₁ = 4

∴ C is (4, 2)

The feasible region satisfying all the conditions is ABCD.

The coordinates of the comer points are A(0, 10), B(1, 5), C(4, 2) and D(12, 0).

Corner points	Z = 3x1 + 2x2
A(0, 10)	20
B(1, 5)	13
C(4, 2)	16
D(12, 0)	36

The minimum value of Z occours at B(1, 5).

∴ The optimal solution is x₁ = 1, x₂ = 5 and Z_min = 13