Question

Maximize Z = 5x + 10y subject to constraints

x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 2y ≤ 10	3x + y ≤ 12
Corresponding equation (of line)	x + 2y = 10	3x + y = 12
Intersection of line with X-axis	(10, 0)	(4, 0)
Intersection of line with Y-axis	(0, 5)	(0, 12)
Region	Origin side	Origin side

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(4, 0), B and C(0, 5).

B is the point of intersection of the lines x + 2y = 10 and 3x + y = 12.

Solving the above equations, we get

x = `14/5` y = `18/5`

Here, the objective function is

Z = 5x + 10y

∴ Z at O(0, 0) = 5(0) + 10(0)

= 0

Z at A(4, 0) = 5(4) + 10(0)

= 20

Z at B`(14/5, 18/5) = 5(14/5) + 10(18/5)`

= 14 + 36

= 50

Z at C(0, 5) = 5(0) + 10(5)

= 50

The maximum value of Z is 50 and it occurs at every point lying on the line segment joining B`(14/5, 18/5)` and C(0, 5).

Hence, there are infinitely many optimal solutions.