Question

Solve the following LPP by graphical method:

Maximize z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.

Answer 1

First we draw the lines AB, AD whose equations are 3x + 2y = 12 and x + y = 4 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + 2y = 12	A(4, 0)	B(0, 6)	≤	origin side of the line AB
AC	x + y = 4	A(4, 0)	C(0, 4)	≥	non-origin side of line AC

The feasible region is the Δ ABC which is shaded in the graph.

The vertices of the feasible region (i.e. corner points) are A(4, 0), B (0, 6) and C (0, 4).

The values of the objective function z = 4x + 6y at these vertices are

z(a) = 4(4) + 6(0) = 16 + 0 = 16

z(B) = 4(0) + 6(6) = 0 + 36 = 36

z(C) = 4(0) + 6(4) = 0 + 24 = 24

∴ z has maximum value 36, when x = 0 and y = 6.