Question

Maximize z = 10x + 25y subject to x + y ≤ 5, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3

Answer 1

To draw the feasible region, construct table as follows:

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

x ≥ 0, y ≥ 0 represent 1^st quadrant.

Shaded portion OABCD is the feasible region,

whose vertices are O(0, 0), A(3, 0), B, C and D(0, 3).

B is the point of intersection of the lines x = 3 and x + y = 5.

Substituting x = 3 in x + y = 5, we get

y = 2

∴ B ≡ (3, 2)

C is the point of intersection of the lines y = 3 and x + y = 5.

Substituting y = 3 in x + y = 5, we get

x = 2

∴ C ≡ (2, 3)

Here, the objective function is Z = 10x + 25y

∴ Z at O(0, 0) = 10(0) + 25(0) = 0

Z at A(3, 0) = 10(3) + 25(0) = 30

Z at B(3, 2) = 10(3) + 25(2) = 30 + 50 = 80

Z at C(2, 3) = 10(2) + 25(3) = 20 + 75 = 95

Z at D(0, 3) = 10(0) + 25(3) = 75

∴ Z has maximum value 95 at C(2, 3).

∴ Z has maximum value 95 when x = 2 and y = 3.