Question

 Solve the following LPP:  

Minimize Z= 120x1+60x2 

Subject to:  

20x1+30x2>= 900  

40x1+30x2>=1200  

x1,x2>=0

Answer 1

Solution:

Minimize $Z= 120x_1+60x_2$

Subject to:  

$20x_1+30x_2\ge 900 \\40x_1+30x_2\ge1200 \\x_1,x_2\ge0$

A(0,40), B(15,20), C(45,0) are corner points.

At A(0,40) , $Z= 120x_1+60x_2 =120(0)+60(40)=2400$ (Minimum)

At B(15,20),$Z= 120x_1+60x_2 =120(15)+60(20)=3000$

At C(45,0),$Z= 120x_1+60x_2 =120(45)+60(0)=5400$

Thus, minimum is 2400 at $x_1=0, x_2=40$