Question

Solve the following linear programming problem graphically.

Maximize Z = 3x₁ + 5x₂ subject to the constraints: x₁ + x₂ ≤ 6, x₁ ≤ 4; x₂ ≤ 5, 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 = 6

x1	0	6
x2	6	0

x₁ = 4 is a line parallel to x₂-axis at a distance of 4 units.

x₂ = 5 is a line parallel to x₁-axis at a distance of 5 units.

The feasible region is OABCD and its co-ordinates are O(0, 0) A(4, 0) D(5, 0) and B is the point of intersection of the lines x₁ + x₂ = 6 and x₁ = 4 

Also C is the point of intersection of the lines x₁ + x₂ = 6 and x₂ = 5

Verification of B:

x₁ + x₂ = 6 and x₁ = 4

4 + x₂ = 6

x₂ = 2

∴ B is (4, 2)

Verification of C:

x₁ + x₂ = 6 and x₂ = 5

x₁ + 5 = 6

x₁ = 1

∴ C is (1, 5)

Corner points	Z = 3x1 + 5x2
O(0, 0)	0
A(4, 0)	12
B(4, 2)	22
C(1, 5)	26
D(5, 0)	15

Maximum of Z occurs at C(1, 5)

∴ The solution is x₁ = 1, x₂ = 5 and Z_max = 26.