Question

Solve the following LPP:

Maximize z = 5x₁ + 6x₂ subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x₁ ≥ 0, x₂ ≥ 0.

Answer 1

First we draw the lines AB and CD whose equations are 2x₁ + 3x₂ = 18 and 2x₁ + x₂ = 12 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	2x1 + 3x2 = 18	A(9, 0)	B(0, 6)	≤	origin side of line AB
CD	2x1 + x2 = 12	C(6, 0)	O(0, 12)	≤	origin side of line CD

The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O (0, 0), C (6, 0), P and B (0, 6).

P is the point of intersection of the lines

2x₁ + 3x₂ = 18        ....(1)

and 2x₁ + x₂ = 12      ....(2)

On subtracting, we get

2x₂ = 6

∴ x₂ = 3

Substituting x₂ = 3 in (2), we get

2x₁ + 3 = 12

∴ x₁ = 9

∴ P is `(9/2, 3)`

The values of objective function z = 5x₁ + 6x₂ at these vertices are

z(O) = 5(0) + 6(0) = 0 + 0 = 0

z(C) = 5(6) + 6(0) = 30 + 0 = 30

z(P) = `5(9/2) + 6(3) = 45/2 + 18 = (45 + 36)/2 = 81/2 = 40.5`

z(B) = 5(0) + 6(3) = 0 + 18 = 18

Maximum value of z is 40.5 when x₁ = `9/2` y₁ = 3.