Solution:

It is given that:

In the figure given, is the feasible region which is open unbounded.

Here, we get

and (ii)

On solving equations equation(i) and equation(ii), we obtain

and

Therefore, the corner points are and

On evaluating the value of , we get

The corner points

Now, the minimum value of is 3 at but as, the feasible region is open bounded so it can or can not be the minimum value of . As a result, in order to face such a situation, we usually draw a graph of and check whether the resulting open half plane has no point in common with feasible region. Otherwise will have no minimum value. Therefore, from the graph, it can be concluded that there is no common point with the feasible region.

Thus, the function has the minimum value at