Question

Maximize z = 5x + 2y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

Answer 1

To draw the feasible region, construct table as follows:

Inequality	3x + 5y ≤ 15	5x + 2y ≤ 10
Corresponding equation (of line)	3x + 5y = 15	5x + 2y = 10
Intersection of line with X-axis	(5, 0)	(2, 0)
Intersection of line with Y-axis	(0, 3)	(0, 5)
Region	Origin side	Origin side

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Here, the objective function is Z = 5x + 2y

∴ Z at O(0, 0) = 5(0) + 2(0) = 0

Z at Q(2, 0) = 5(2) + 2(0) = 10

Z at R `(20/19, 45/19) = 5 (20/19) + 2(45/19) = 10`

Z at A(0, 3) = 5(0) + 2(3) = 6

 The maximum value of Z is 10 and it occurs at every point lying on the line segment joining `"R"(20/19, 45/19)` and Q(2, 0).

Hence, there are infinitely many optimal solutions.