Solution:

It is given that: $Z=3 x+4 y$ and the constraints $x+y \leq 1, x \geq 0$ $\mathrm{y} \geq 0$

Taking $x+y=1$, we have

$$\begin{tabular}{|l|l|l|}
\hline$x$ & 1 & 0 \\
\hline$y$ & 0 & 1 \\
\hline
\end{tabular}$$

Now, plotting all the constrain equations it can be seen that the shaded area $\mathrm{OAB}$ is the feasible region determined by the constraints.

The area is feasible. Therefore, maximum value will occur at the corner points $\mathrm{O}(0,0), \mathrm{A}(1,0), \mathrm{B}(0,1)$

On evaluating the value of $\mathrm{Z}$, we obtain

$$\begin{tabular}{|l|l|}
\hline Corner points & Value of Z \\
\hline $0(0,0)$ & $3(0)+4(0)=0$ \\
\hline $\mathrm{A}(1,0)$ & $3(1)+4(0)=3$ \\
\hline $\mathrm{B}(0,1)$ & $3(0)+4(1)=4$
\hline
\end{tabular}$$

It can be seemn from the above table that the maximum value of $Z$ is $4 .$

So, the maximum value of $\mathrm{Z}$ is 4 at $(0,1)$.