Question

Solve the Linear Programming problem graphically:

Maximize z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find the maximum value of z.

Answer 1

To draw the feasible region, construct table as follows:

Inequality	x + 4y ≤ 24	3x + y ≤ 21	x + y ≤ 9
Corresponding equation (of line)	x + 4y = 24	3x + y = 21	x + y = 9
Intersection of line with X-axis	(24, 0)	(7, 0)	(9, 0)
Intersection of line with Y-axis	(0, 6)	(0, 21)	(0, 9)
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(7, 0), B, C and D(0, 6).

B is the point of intersection of the lines 3x + y = 21 and x + y = 9.

Solving the above equations, we get

x = 6, y = 3

∴ B ≡ (6, 3)

C is the point of intersection of the lines x + 4y = 24 and x + y = 9.

Solving the above equations, we get

x = 4, y = 5

∴ C ≡ (4, 5)

Here, the objective function is Z = 3x + 5y

∴ Z at O(0, 0) = 3(0) + 5(0) = 0

Z at A(7, 0) = 3(7) + 5(0) = 21

Z at B(6, 3) = 3(6) + 5(3) = 18 + 15 = 33

Z at C(4, 5) = 3(4) + 5(5) = 12 + 25 = 37

Z at D(0, 6) = 3(0) + 5(6) = 30

∴ Z has maximum value 37 at C(4, 5).

∴ Z has maximum value 37 when x = 4, y = 5.