In: Advanced Math
a)
let x1 be the oats and x2 be rice.
Lp problem
MIN z = 0.05x1 + 0.03x2
subject to
8x1 + 6x2 >= 48
x1 + 2x2 >= 12
and x1,x2 >= 0
b)
1. To draw constraint 8x1+6x2≥48→(1)
Treat it as 8x1+6x2=48
When x1=0 then x2=?
⇒8(0)+6x2=48
⇒6x2=48
⇒x2=486=8
When x2=0 then x1=?
⇒8x1+6(0)=48
⇒8x1=48
⇒x1=488=6
x1 | 0 | 6 |
x2 | 8 | 0 |
2. To draw constraint x1+2x2≥12→(2)
Treat it as x1+2x2=12
When x1=0 then x2=?
⇒(0)+2x2=12
⇒2x2=12
⇒x2=122=6
When x2=0 then x1=?
⇒x1+2(0)=12
⇒x1=12
x1 | 0 | 12 |
x2 | 6 | 0 |
Graphical solution
The minimum value of the objective function z=0.24 occurs at the
extreme point (0,8).
Hence, the optimal solution to the given LP problem is : x1=0,x2=8
and min z=0.24.
optimal solution = (2.4 , 4.8)
c)
Hence, the optimal solution to the given LP problem is : x1=2.4,x2=4.8 and min z=0.41.
There is no effect on the optimal solution. Cost will increase.