Linear Programming — Class 12 Maths Solution

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 the following conditions as per the question, $Z = 4x + y$.

[since, $x + 2y = 4$ and $x + y = 3 \Rightarrow y = 1$ and $x = 2$]

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).