Question

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

Maximize, z = 9x + 13y

Subject to, 2x + 3y ≤ 18,

                  2x + y ≤ 10

                  x ≥ 0, y ≥ 0

Answer 1

First, we draw the lines AB and CD whose equations are 2x + 3y = 18 and 2x + y = 10, respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	2x + 3y = 18	A (9, 0)	B (0, 6)	≤	origin side of the line AB
CD	2x + y = 10	C (5, 0)	D (0, 10)	≤	origin side of the line CD

The feasible region is OCPBO which is shaded in the graph.

The vertices of the feasible region are O(0, 0), C(5, 0), P and B(0, 6).

P is the point of intersection of the lines

2x + 3y = 18

And 2x + y = 10

On subtracting, we get

2y = 8

∴ y = 4

Substituting y = 4 in 2x + y = 10, we get

2x + 4 = 10

∴ 2x = 6

∴ x = 3

∴ P ≡ (3, 4)

∴ The corner points of the feasible region are O(0, 0), C(5, 0), P(3, 4) and B(0, 6).

The values of the objective function z = 9x + 13y at these corner points are

z(O) = 9(0) + 13(0)

= 0 + 0

= 0

z(C) = 9(5) + 13(0)

= 45 + 0

= 45

z(P) = 9(3) + 13(4)

= 27 + 52

= 79

z(B) = 9(0) + 13(6)

= 0 + 78

= 78

∴ z has maximum value 79, when x = 3 and y = 4.