Question

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

Minimize: z = 8x + 10y

Subject to: 2x + y ≥ 7, 2x + 3y ≥ 15, y ≥ 2, x ≥ 0, y ≥ 0.

Answer 1

First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	2x + y = 7	A(3.5, 0)	B(0, 7)	≥	non-origin side of line AB
CD	2x + 3y = 15	C(7.5, 0)	D(0, 5)	≥	non-origin side of line CD
EF	y = 2	–	F(0, 2)	≥	non-origin side of line EF

The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0, 7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.

Substituting y = 2 in 2x + 3y = 15, we get

2x + 3(2) = 15

∴ 2x = 9

∴ x = 4.5 

∴ P = (4.5, 2)

Q is the point of intersection of the lines

2x + 3y = 15         .....(1)

and 2x + y = 7      .....(2)

On subtracting, we get

2y = 8

∴ y = 4

∴ from (2), 2x + 4 = 7

∴ 2x = 3

∴ x = 1.5

∴ Q = (1.5, 4)

The values of the objective function z = 8x + 10y at these vertices are

z(P) = 8(4.5) + 10(2)

= 36 + 20 

= 56

z(Q) = 8(1.5) + 10(4)

= 12 + 40 

= 52

z(B) = 8(0) + 10(7)

= 0 + 70

= 70

∴ z has minimum value 52, when x = 1.5 and y = 4.