Question

Solve the following linear programming problem graphically.

Minimize Z = 200x₁ + 500x₂ subject to the constraints: x₁ + 2x₂ ≥ 10; 3x₁ + 4x₂ ≤ 24 and x₁ ≥ 0, 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 + 2x₂ = 10

x1	0	10
x2	5	0

3x₁ + 4x₂ = 24

x1	0	8
x2	6	0

The feasible region is ABC and its co-ordinates are A(0, 5) C(0, 6) and B is the point of intersection of the lines

x₁ + 2x₂ = 10 ..........(1)

3x₁ + 4x₂ = 24 .........(2)

Verification of B:

3x₁ + 6x₂ = 30 ..........[(1) × 3]
3x₁ + 4x₂ = 24 .........(2)
−     −       −       
2x₂ = 6

x₂ = 3

From (1), x₁ + 6 = 10

x₁ = 4

∴ B is (4, 3)

Corner points	Z = 200x1 + 500x2
A(0, 5)	2500
B(4, 3)	2300
C(0, 6)	3000

Minimum value occurs at B(4, 3)

∴ The solution is x₁ = 4, x₂ = 3 and Z_min = 2300.