The feasible region for a LPP is shown in the following figure. Evaluate Z=4x +y at each of the corner points of this region. Find the minimum value of Z, if it exists.
The feasible region for a LPP is shown in the following figure. Evaluate Z=4x +y at each of the corner points of this region. Find the minimum value of Z, if it exists.
From the shaded region, it is clear that feasible region is unbounded with the corner points A(4,0), B(2,1) and C (0,3)
Also, we have Z =4x + y.
[since, x +2y =4 and x+y=3⇒y=1 and x =2]
Corner pointsCorresponding value of Z(4,0)16(2,1)9(0,3)3←Minimum
Now, we see that 3 is the smallest value of Z at the corner point (0,3). Note that here we see that, the region is unbounded, therefore 3 may or may not be the minimum value of Z.
To decide this issue, we graph the inequality 4x+y < 3 and check whether the resulting open half plan has no point in common with feasible region otherwise, Z has no minimum value.
From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0,3).
From the shaded region, it is clear that feasible region is unbounded with the corner points A(4,0), B(2,1) and C (0,3)
Also, we have Z =4x + y.
[since, x +2y =4 and x+y=3⇒y=1 and x =2]
Corner pointsCorresponding value of Z(4,0)16(2,1)9(0,3)3←Minimum
Now, we see that 3 is the smallest value of Z at the corner point (0,3). Note that here we see that, the region is unbounded, therefore 3 may or may not be the minimum value of Z.
To decide this issue, we graph the inequality 4x+y < 3 and check whether the resulting open half plan has no point in common with feasible region otherwise, Z has no minimum value.
From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0,3).
