Given objective function is
Z=x+2y
We have to minimize Z on constraints
2x+y≥3
x+2y≥6
x≥0,y≥0
After plotting the inequalities we got the feasible region as shown in the image
Now there are two corner points (0,3) and (6,0) lying on same line x+2y=6
Value at corner points are :
Corner Points | Value of Z=x+2y |
(0,3) | 6 (minimum) |
(6,0) | 6 (minimum) |
Since, feasible region is unbounded. So, 6 may or may not be minimum value.
Now to check if 6 is minimum or not, we have to draw Z<6⇒x+2y<6
Since this region doesn't have any common region with feasible region.
So, 6 is the minimum value of Z.