Question

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

Minimize : Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.

Answer 1

To draw the feasible region, construct table as follows:

Inequality	5x + y ≥ 5	x + y ≥ 3
Corresponding equation (of line)	5x + y = 5	x + y = 3
Intersection of line with X-axis	(1, 0)	(3, 0)
Intersection of line with Y-axis	(0, 5)	(0, 3)
Region	Non-origin side	Non-origin side

Shaded portion XABCY is the feasible region,
whose vertices are A(3, 0), B and C (0, 5).
B is the point of intersection of the lines x + y = 3 and 5x + y = 5
Solving the above equations, we get

x = `(1)/(2), y = (5)/(2)`

∴ B ≡ `(1/2, 5/2)`
Here, the objective function is Z = 7x + y
Z at A(3, 0) = 7(3) + 0 = 21

Z at B`(1/2, 5/2) = 7(1/2) + (5)/(2)`

= `(7)/(2) + (5)/(2)` = 6
Z at C (0, 5) = 7(0) + 5 = 5
∴ Z has minimum value 5 at C(0, 5).
∴ Z is minimum, when x = 0 and y = 5.