Question

Solve the following L.P.P. : 
Minimize : Z = 4x + 10y, 
Subject to : 2x + 5y ≤10 , 5x + 3y ≤ 15, 
                  x + 2y  ≥ 30, x ≥ 0, y ≥ 0. 

Answer 1

2x+ 5y= 10	x	y	Point (x, y)
	5	0	A (5, 0)
	0	2	B (0, 2)

5x+3y= 15	x	y	Point (x, y)
	3	0	C (3, 0)
	0	5	D (0, 5)

The shaded region OCPB is the feasible region. Since P is the point of intersection of two lines
2x+ 5y = 10 and 5x+ 3y = 15. 
∴ Solving these two equations we get 
P = `(45/19 , 20/19)`

Corner Points	Value of Z = 4x+ 10y
O (0, 0)	Zc = 0
C (3, 0)	Zc = 4(3) + 10(0) = 12
P`(45/19 , 20/19)`	ZP = `4(45/19) + 10(20/19) = 20`
B(0 , 2)	ZB =  4(0) + 10(2) = 20

∴ Z_max = 20 at points P and B. 

As P and B are adjacent vertices of the feasible region. 

Z_max = 20 at each point of segment PB. 

∴ The L.P.P. has infinite number of solutions .