Example 13 Find the maximum value of the function $f(x)=\sqrt{10 x-9-x^{2}}+\sqrt{68 x-256-x^{2}}$.
Solution 1: From $10 x-9-x^{2} \geqslant 0$ and $68 x-256-x^{2} \geqslant 0$, we find the domain to be $4 \leqslant x \leqslant 9$.
$$
\text{Since } \begin{aligned}
f(x) & =\sqrt{10 x-9-x^{2}}+\sqrt{68 x-256-x^{2}} \\
& =\sqrt{(x-1)(9-x)}+\sqrt{(64-x)(x-4)}, \\
\text{and }(\sqrt{x-1})^{2} & +(\sqrt{64-x})^{2}=63,(\sqrt{9-x})^{2}+(\sqrt{x-4})^{2}=5 .
\end{aligned}
$$

We can let $\sqrt{x-1}=\sqrt{63} \cos \alpha, \alpha \in\left[0, \frac{\pi}{2}\right]$,
$$
\sqrt{9-x}=\sqrt{5} \cos \beta, \beta \in\left[0, \frac{\pi}{2}\right] .
$$

Thus, $f(x)=g(\alpha, \beta)=\sqrt{63} \cdot \sqrt{5}(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta)=\sqrt{35} \cos (\alpha-\beta)$. When $\alpha=\beta$, i.e., $\frac{x-1}{63}=\frac{9-x}{5}$, i.e., $x=\frac{143}{17} \in[4,9]$,
$f(x)$ reaches its maximum value of $3 \sqrt{35}$.
Solution 2: Construct a right trapezoid $A B C D$ such that $A B=\sqrt{x-1}, C D=\sqrt{x-4}, B C=B E+E C=$ $\sqrt{64-x}+\sqrt{9-x}$, as shown in Figure 5-6.
From $S_{\text{trapezoid } \triangle B C D}=S_{\triangle A B E}+S_{\triangle C D E}+S_{\triangle A E D}$, we know
$$
\begin{array}{l}
\frac{1}{2}(\sqrt{x-1}+\sqrt{x-4}) \cdot(\sqrt{64-x}+\sqrt{9-x}) \\
=\frac{1}{2} \sqrt{x-1} \cdot \sqrt{64-x}+\frac{1}{2} \sqrt{9-x} \cdot \sqrt{x-4}+\frac{1}{2} \sqrt{5} \cdot \sqrt{63} \cdot \sin \angle A E D .
\end{array}
$$

Thus, $f(x)=\sqrt{x-1} \cdot \sqrt{9-x}+\sqrt{64-x} \cdot \sqrt{x-4} \leqslant \sqrt{63} \cdot \sqrt{5}=3 \sqrt{35}$ is the desired maximum.
Solution 3: By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
f(x) & =\sqrt{x-1} \cdot \sqrt{9-x}+\sqrt{64-x} \cdot \sqrt{x-4} \\
& \leqslant \sqrt{(x-1+64-x) \cdot(9-x+x-4)}=\sqrt{63 \cdot 5}=3 \sqrt{35},
\end{aligned}
$$

where equality holds if and only if $\frac{x-1}{9-x}=\frac{64-x}{x-4}$, i.e., $x=\frac{143}{17} \in[4,9]$. Therefore, the maximum value of $f(x)$ is $3 \sqrt{35}$.
Using the Cauchy-Schwarz inequality method can also be seen as a vector method: $\boldsymbol{u} \cdot \boldsymbol{v} \leqslant|\boldsymbol{u}| \cdot|\boldsymbol{v}|$.