Solution:
It is given that: 
![\[\begin{tabular}{|l|l|l|} \hline$x$ & 4 & 3 \\ \hline $\mathrm{v}$ & 3 & 4 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a89c1e28a02ef115f82e3f5e78e95c4b_l3.png "Rendered by QuickLaTeX.com")
And, taking 
![\[\begin{tabular}{|l|l|l|} \hline$x$ & 1 & $-3$ \\ \hline$y$ & 2 & 0 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc615c7ae759639a0eb48710155bac4b_l3.png "Rendered by QuickLaTeX.com")
Now, plotting all the constrain equations it can be seen that the shaded area 
The feasible reqion is bounded with four corners 
Therefore, the maximum value can occur at any corner.
On evaluating the value of 
![\[\begin{tabular}{|l|l|} \hline Corner points & Value of $\mathrm{Z}$ \\ \hline $\mathrm{O}(0,0)$ & $13(0)-15(0)=0$ \\ \hline $\mathrm{A}(7,0)$ & $13(7)-15(0)=91$ \\ \hline $\mathrm{B}(3,4)$ & $13(3)-15(4)=-21$ \\ \hline $\mathrm{C}(0,2)$ & $13(0)-15(2)=-30$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-65eeddd2f7496352bd37c796573f6a2c_l3.png "Rendered by QuickLaTeX.com")
It can be seen from the above table it’s seen that the minimum value of 
