(11) (15 points) Find the maximum and minimum values of the function $y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$.
(11) The domain of the function is $[0,13]$. Because
$$
\begin{aligned}
y & =\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
& =\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
& \geqslant \sqrt{27}+\sqrt{13} \\
& =3 \sqrt{3}+\sqrt{13},
\end{aligned}
$$

when $x=0$, the equality holds. Therefore, the minimum value of $y$ is $3 \sqrt{3}+\sqrt{13}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
y^{2} & =(\sqrt{x}+\sqrt{x+27}+\sqrt{13-x})^{2} \\
& \leqslant\left(\frac{1}{2}+1+\frac{1}{3}\right)[2 x+(x+27)+3(13-x)] \\
& =121,
\end{aligned}
$$

so $y \leqslant 11$.
By the condition for equality in the Cauchy-Schwarz inequality, we get $4 x=9(13-x)=x+27$, solving which gives $x=9$. Therefore, when $x=9$, the equality holds. Thus, the maximum value of $y$ is 11.