Example 7 Find the maximum and minimum values of the function $y=2 \sqrt{x+5}+\sqrt{4-x}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Let $\sqrt{x+5}=u, \sqrt{4-x}=v(u \geqslant 0, v \geqslant 0)$,
then $\left\{\begin{array}{l}u^{2}+v^{2}=9, \\ y=2 u+v .\end{array}\right.$
Substituting $v=y-2 u$ into
$u^{2}+v^{2}=9$ gives
$5 u^{2}-4 y u+y^{2}-9=0$,
Given $u \geqslant 0, v \geqslant 0$, we have
$$
\left\{\begin{array}{l}
-4 y<0, \\
y^{2}-9 \geqslant 0, \\
\Delta=16 y^{2}-5 \cdot 4\left(y^{2}-9\right) \geqslant 0 .
\end{array}\right.
$$

Thus $3 \leqslant y \leqslant 3 \sqrt{5}$,
If $0<y<3$, it is easy to see that $u^{2}+v^{2}<(2 u+v)^{2}=y^{2}<9$, which does not meet the condition.
Therefore, $y$ has a maximum value of $3 \sqrt{5}$ and a minimum value of 3.