#### Answer

The maximum value of the given function is $26$.

#### Work Step by Step

Let us consider the inequalities and substitute the equals symbol in place of the inequality to rewrite the equation as given below:
Find two solutions of the linear equation to plot the graph of the linear equation $ x+y=6$
To find the value of the x-intercept, put y = 0 as given below:
$\begin{align}
& x+y=6 \\
& x+0=6 \\
& x=6
\end{align}$
To find the value of the y-intercept, substitute x = 0 as follows:
$\begin{align}
& 0+y=6 \\
& y=6
\end{align}$
Therefore, plot the intercepts $\left( 6,0 \right)\text{ and }\left( 0,6 \right)$ and draw a solid line passing through these points because the inequality contains the $\le $ symbol in which the equality is included.
Now, this solid line divides the plane into three regions: the line itself and two half planes.
Then, take origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade:
$\begin{align}
x+y\le 6 & \\
0+0\overset{?}{\mathop{\le }}\,6 & \\
0\le 6 & \\
\end{align}$
Since the test point satisfies the inequality, shade the half plane containing this point, that is, the origin $\left( 0,0 \right)$.
Now, to plot the graph of the inequality $ x\ge 0$, graph the line $ x=0$ which is a vertical line parallel to the y-axis and shade the right part as the inequality contains the $\ge $ symbol. Similarly, to plot the graph of the inequality $ y\ge 0$, graph the line $ y=0$ which is a horizontal line parallel to the x-axis and shade the upper part as the inequality contains the $\ge $ symbol.
Similarly, to plot the graph of $ x\ge 2$, graph the line $ x=2$ and shade the right part as the inequality contains the $\ge $ symbol.
So, the graph of these inequalities is the common shaded region.
Substitute the value of $ x=2$ in the equation $ x+y=6$ to obtain the value of the y-intercept:
$\begin{align}
& 2+y=6 \\
& y=6-2 \\
& y=4
\end{align}$
Therefore, the intersection point of $ x+y=6$ and $ x=2$ is $\left( 2,4 \right)$.
And the corner points of the given graph are $(2,0),\ (6,0),\ \text{ and }\ \left( 2,4 \right)$.
The value of the objective function at each corner point can be computed as given below:
$\begin{matrix}
\text{Corner points} & \begin{align}
& \text{Value of} \\
& z=3x+5y \\
\end{align} & {} \\
\left( 2,0 \right) & 3\left( 2 \right)+5\left( 0 \right)=6 & {} \\
\left( 6,0 \right) & 3\left( 6 \right)+5\left( 0 \right)=18 & {} \\
\left( 2,4 \right) & 3\left( 2 \right)+5\left( 4 \right)=26 & \text{Max} \\
\end{matrix}$
Thus, the maximum value of the objective function is $26$ at the corner point $\left( 2,4 \right)$.