(4) The maximum value of the function $f(x)=\sqrt{x-5}+\sqrt{24-3 x}$ is $(\quad)$.
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) $3 \sqrt{3}$
(4) $\mathrm{C}$ Solution 1 The domain of $f(x)$ is $5 \leqslant x \leqslant 8$, and
$$
\begin{aligned}
f^{\prime}(x) & =\frac{1}{2 \sqrt{x-5}}+\frac{-3}{2 \sqrt{24-3 x}} \\
& =\frac{\sqrt{24-3 x}-3 \sqrt{x-5}}{2 \sqrt{x-5} \cdot \sqrt{24-3 x}}=0,
\end{aligned}
$$

Solving this gives $x=\frac{23}{4}$. Since $f(5)=3, f\left(\frac{23}{4}\right)=2 \sqrt{3}, f(8)=\sqrt{3}$, we have
$$
f(x)_{\max }=f\left(\frac{23}{4}\right)=2 \sqrt{3} .
$$

Solution 2 The domain of $f(x)$ is $5 \leqslant x \leqslant 8$,
$$
\begin{aligned}
f^{2}(x) & =(1 \cdot \sqrt{x-5}+\sqrt{3} \cdot \sqrt{8-x})^{2} \\
& \leqslant(1+3)(x-5+8-x)=12,
\end{aligned}
$$

When and only when $\frac{\sqrt{x-5}}{1}=\frac{\sqrt{8-x}}{\sqrt{3}}$, i.e., $x=\frac{23}{4}$, $f(x)$ reaches its maximum value $2 \sqrt{3}$.