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

To draw the feasible region, construct table as follows:

Inequality	x ≤ 4	y ≤ 6	x + y ≤ 6
Corresponding equation (of line)	x = 4	y = 6	x + y = 6
Intersection of line with X-axis	(4, 0)	–	(6, 0)
Intersection of line with Y-axis	–	(0, 6)	(0, 6)
Region	Origin side	Origin side	Origin side

Shaded portion OABC is the feasible region,
whose vertices are O(0, 0), A(4, 0), B and C(0, 6).
B is the point of intersection of the lines x = 4 and x + y = 6

Substituting x = 4 in x + y = 6, we get 
y = 2
∴ B ≡ (4, 2)
Here, the objective function is Z = 11x + 8y
∴ Z at O(0, 0) = 11(0) + 8(0) = 0
Z at A(4, 0) = 11(4) + 8(0) = 44
Z at B(4, 2) = 11(4) + 8(2) = 44 + 16 = 60
Z at C(0, 6) = 11(0) + 8(6) = 48
∴ Z has maximum value 60 at B(4, 2)
∴ Z is maximum, when x = 4 and y = 2.