#### Answer

The value of the objective function at the corner point $\left( 2,2 \right)$ is $10$, at the point $\left( 4,0 \right)$ is $8$, at the point $\left( \frac{1}{2},\frac{1}{2} \right)$ is $\frac{5}{2}$ and at the point $\left( 1,0 \right)$ is $2$.
The maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$ and the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$.

#### Work Step by Step

The corner points of the given graph are $\left( 2,2 \right)$, $\left( 4,0 \right)$, $\left( \frac{1}{2},\frac{1}{2} \right)$, $\left( 1,0 \right)$ and the objective function is $ z=2x+3y $.
The value of objective function at each corner point can be computed as given below.
At the point $\left( 1,0 \right)$:
$\begin{align}
& \left( 1,0 \right)=2\times 1+3\times 0 \\
& =2
\end{align}$
At the point $\left( \frac{1}{2},\frac{1}{2} \right)$:
$\begin{align}
& z=2\times \frac{1}{2}+3\times \frac{1}{2} \\
& =\frac{5}{2}
\end{align}$
Next, at the point $\left( 4,0 \right)$:
$\begin{align}
& z=2\times 4+3\times 0 \\
& =8
\end{align}$
Also, at the point $\left( 2,2 \right)$:
$\begin{align}
& z=2\times 2+3\times 2 \\
& =10
\end{align}$
Therefore, the maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$,
And, the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$.
Thus, the value of the objective function at the point $\left( 2,2 \right)$ is $10$, at the point $\left( 4,0 \right)$ is $8$, at the point $\left( \frac{1}{2},\frac{1}{2} \right)$ is $\frac{5}{2}$, and at the point $\left( 1,0 \right)$ is $2$.
The maximum value of the objective function is $10$ at the point $\left( 2,2 \right)$ and the minimum value of the objective function is $2$ at the point $\left( 1,0 \right)$.