Solution:
It is given that: 
Now construct a constrain table for the above, we have
Here, it can be seen that 
Table for 
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 8 \\ \hline $\mathrm{y}$ & 2 & 0 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-df44431c49c119fb39d011eb16c6eb92_l3.png "Rendered by QuickLaTeX.com")
Table for 
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 6 \\ \hline $\mathrm{y}$ & 4 & 0 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e9aa79443edf0df7d2bdd8d2069a8960_l3.png "Rendered by QuickLaTeX.com")
Table for 
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 3 & 0 \\ \hline $\mathrm{y}$ & 0 & 9 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-477338485fd6be6f7deae8264267cad3_l3.png "Rendered by QuickLaTeX.com")
On solving equations 
Here, it can be seen that 
Let’s evaluate the value of 
![\[\begin{tabular}{|l|l|} \hline Corner points & Value of $Z=x+y$ \\ \hline $0(0,0)$ & $Z=0+0=0$ \\ \hline$A(3,0)$ & $Z=3+0=3$ \\ \hline$B(28 / 11,15 / 11)$ & $Z=28 / 11+15 / 11=43 / 11=3.9$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c61b267a99779f6eab274c06646b199e_l3.png "Rendered by QuickLaTeX.com")
It is observed from the above table that the maximum value of 
Thus, the maximum value of 
