Question

Maximize Z = 400x + 500y subject to constraints

x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 2y ≤ 80	2x + y ≤ 90
Corresponding equation (of line)	x + 2y = 80	2x + y = 90
Intersection of line with X-axis	(80, 0)	(45, 0)
Intersection of line with Y-axis	(0, 40)	(0, 90)
Region	Origin side	Origin side

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(45, 0), B and C(0, 40).

B is the point of intersection of the lines 2x + y = 90 and x + 2y = 80.

Solving the above equations, we get

x = `100/3`, y = `70/3`

∴ B = `(100/3, 70/3)`

Here, the objective function is

Z = 400x + 500y

∴ Z at O(0, 0) = 400(0) + 500(0) = 0

Z at A(45, 0) = 400(45) + 500(0) = 18000

X at B`(100/3, 70/3) = 400(100/3) + 500(70/3)`

= `40000/3 + 35000/3`

= `75000/3`

= 25000

Z at C(0, 40) = 400(0) + 500(40)

= 20000

∴ Z has maximum value 25000 at x = `100/3` and y = `70/3`.