Question

Solve the following L.P.P. by graphical method:

Maximize: Z = 4x + 6y

Subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.

Answer 1

The draw the feasible region, construct table as follows:

Inequality	3x + 2y ≤ 12	x + y ≥ 4
Corresponding equation (of line)	3x + 2y = 12	x + y = 4
Intersection of line with X-axis	(4, 0)	(4, 0)
Intersection of line with Y-axis	(0, 6)	(0, 4)
Region	Origin side	Non-origin side

Shaded portion ABC is the feasible region,

Whose vertices are A(4, 0), B(0, 6), C(0, 4).

Here, the objective function is Z = 4x + 6y

∴ Z at A(4, 0) = 4(4) + 6(0) = 16

Z at B(0, 6) = 4(0) + 6(6) = 36

Z at C(0, 4) = 4(0) + 6(4) = 24

∴ Z has maximum value 36 at B(0, 6)

∴ Z is maximum when x = 0 and y = 6.