1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$ .
$-1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$

Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \times \frac{13-x}{4}+3 \times \frac{x}{9}\right)} \\
=11 .
\end{array}
$$

When $x=9$, the equality holds.
Therefore, the maximum value of $f(x)$ is 11.