Question

Solve the following LPP by graphical method:

Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0

Answer 1

First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	x = 4	A(4, 0)	-	≤	origin side of the line AB
CD	y = 6	-	D(0, 6)	≤	origin side of the line CD
EF	x + y = 6	E(6, 0)	D(0, 6)	≤	origin side of the line ED

The feasible region is the shaded portion OAPDO in the graph.

The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)

P is point of intersection of lines x + y = 6 and x = 4.

Substituting x = 4 in x + y = 6, we get

4 + y = 6    ∴ y = 2          ∴ P is (4, 2)

∴ the corner points of feasible region are O (0, 0), A (4, 0), P (4, 2) and D (0, 6).

The values of the objective function z = 11x + 8y at these vertices are

z(O) = 11(0) + 8(0) = 0 + 0 = 0

z(a) = 11(4) + 8(0) = 44 + 0 = 44

z(P) = 11(4) + 8(2) = 44 + 16 = 60

z(D) = 11(0) + 8(6) = 48

∴ z has maximum value 60, when x = 4 and y = 2.