Answer
The maximum is $f(2)=4\sqrt{2}$ and the minimum is $f\left( 0\right) =2\sqrt{6}$
Work Step by Step
Given $$y=(2+x) \sqrt{2+(2-x)^{2}}, \quad[0,2]$$
Since
\begin{align*}
\frac{dy}{dx}&=\frac{d}{dx}\left(2+x\right)\sqrt{2+\left(2-x\right)^2}+\frac{d}{dx}\left(\sqrt{2+\left(2-x\right)^2}\right)\left(2+x\right)\\
&=\frac{2x^2-4x+2}{\sqrt{x^2-4x+6}}
\end{align*}
Find critical points
\begin{align*}
f'(x)&=0\\
\frac{2x^2-4x+2}{\sqrt{x^2-4x+6}}&=0\\
2x^2-4x+2&=0\\
2(x-1)^2&=0
\end{align*}
Then $x=1$. Because $1\in [0,2]$, then $f(x)$ has a critical point at $x= 1 $, since
\begin{align*}
f(0)&=2\sqrt{6}\\
f(1)&= 3\sqrt{3}\\
f(2)&=4\sqrt{2}
\end{align*}
Then the maximum is $f(2)=4\sqrt{2}$ and the minimum is $f\left( 0\right) =2\sqrt{6}$
