Question

Maximize Z = 2x + 3y subject to constraints

x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 4y ≤ 8	3x + 2y ≤ 14
Corresponding equation (of line	x + 4y = 8	3x + 2y = 14
Intersection of line with X-axis	(8, 0)	`(14/3, 0)`
Intersection of line with Y-axis	(0, 2)	(0, 7)
Region	Origin side	Origin side

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A`(14/3, 0)`, B and C(0, 2).

B is the point of intersection of the lines x + 4y = 8 and 3x + 2y = 14.

Solving the above equations, we get

x = 4, y = 1

∴ B = (4, 1)

Here, the objective function is

Z = 2x + 3y

∴ Z at O(0, 0) = 2(0) + 3(0) = 0

Z at A`(14/3, 0) = 2(14/3) + 3(0) = 28/3`

Z at B(4, 1) = 2(4) + 3(1) = 8 + 3 = 11

Z at C(0, 2) = 2(0) + 3(2) = 6 

∴ Z has maximum value 11 at x = 4 and y = 1.