Question

Solve the following L.P.P. by graphical method :

Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find 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

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 = 4y = 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 is maximum, when x = 4, y = 5.