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

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

Thus, $y \leqslant 11$.
By the condition for equality in the Cauchy-Schwarz inequality, we get
$$
4 x=9(13-x)=x+27 \Rightarrow x=9 \text {. }
$$

Thus, when $x=9$, the equality in (1) holds.
Therefore, the maximum value of $y$ is 11.