Question

Solve the following LPP:

Maximize z =60x + 50y  subject to

x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0.

Answer 1

We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	x + 2y = 40	A(40,0)	B(0,20)	≤	origin side of line AB
CD	3x + 2y = 60	C(20,0)	D(0,30)	≤	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 (20, 0), P and B (0, 20).
P is the point of intersection of the lines.

3x + 2y = 60       ....(1)

and x + 2y = 40     .....(2)

On subtracting, we get

2x = 20      ∴ x = 10

Substituting x = 10 in (2), we get

10 + 2y = 40

∴ 2y = 30

∴ y = 15

∴ P is (10, 15)

The values of the objective function z = 60x + 50y at these vertices are

z(O) = 60(0) + 50(0) = 0 + 0 = 0

z(C) = 60(20) + 50(0) = 1200 + 0 = 1200

z(P) = 60(10) + 50(15) = 600 + 750 = 1350

z(B) = 60(0) + 50(20) = 0 + 1000 = 1000

∴ z has maximum value 1350 at x = 10, y = 15.