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
![\[\begin{tabular}{|l} $\mathrm{B}(2,1)$ $\mathrm{C}(0,3)$ \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6077b4fe69a90bffde567225b485bf2b_l3.png "Rendered by QuickLaTeX.com")
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