Solution:

It is given that: $Z=x+y$ subject to constraints, $x+4 y \leq 8$

$2 x+3 y \leq 12,3 x+y \leq 9, x \geq 0, y \geq 0$

Now construct a constrain table for the above, we have

Here, it can be seen that $\mathrm{OABC}$ is the feasible region

Table for $x+4 y=8$
$$\begin{tabular}{|l|l|l|}
\hline $\mathrm{x}$ & 0 & 8 \\
\hline $\mathrm{y}$ & 2 & 0 \\
\hline
\end{tabular}$$
Table for $2 x+3 y=12$
$$\begin{tabular}{|l|l|l|}
\hline $\mathrm{x}$ & 0 & 6 \\
\hline $\mathrm{y}$ & 4 & 0 \\
\hline
\end{tabular}$$
Table for $3 x+y=9$
$$\begin{tabular}{|l|l|l|}
\hline $\mathrm{x}$ & 3 & 0 \\
\hline $\mathrm{y}$ & 0 & 9 \\
\hline
\end{tabular}$$
On solving equations $x+4 y £ 8$ and $3 x+y £ 9$, we obtain
$x=28 / 11 \text { and } y=15 / 11$
Here, it can be seen that $\mathrm{OABC}$ is the feasible region whose corner points are $\mathrm{O}(0,0), \mathrm{A}(3,0), \mathrm{B}(28 / 11,15 / 11)$ and $\mathrm{C}(0,2) .$

Let’s evaluate the value of $Z$

$$\begin{tabular}{|l|l|}
\hline Corner points & Value of $Z=x+y$ \\
\hline $0(0,0)$ & $Z=0+0=0$ \\
\hline$A(3,0)$ & $Z=3+0=3$ \\
\hline$B(28 / 11,15 / 11)$ & $Z=28 / 11+15 / 11=43 / 11=3.9$ \\
\hline
\end{tabular}$$

$\mathrm{C}(0,2)$

$\mathrm{Z}=0+2=2$

It is observed from the above table that the maximum value of $\mathrm{Z}$ is $3.9$

Thus, the maximum value of $\mathrm{Z}$ is $3.9$ at $(28 / 11,15 / 11)$.