$$
f(x)=x\left(\sqrt{36-x^{2}}+\sqrt{64-x^{2}}\right)
$$

Find the maximum value of the function.
Algebraic solution Using the Cauchy-Schwarz inequality, we get
$$
\begin{aligned}
f(x) & =x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
& \leqslant \sqrt{\left(x^{2}+36-x^{2}\right)\left(64-x^{2}+x^{2}\right)}=48 .
\end{aligned}
$$

Geometric solution Construct $\triangle A B C, A D \perp B C$, and let $A B=6$, $A C=8, A D=x$. Then
$$
\begin{array}{l}
2 S_{\triangle A B C}=2 S_{\triangle A D B}+2 S_{\triangle A D C} \\
=x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
=6 \times 8 \sin \angle B A C .
\end{array}
$$

Obviously, when
$$
\angle B A C=90^{\circ} \Leftrightarrow A B \perp A C \Leftrightarrow B C=10, A D=\frac{24}{5}
$$

$S_{\triangle A B C}$ achieves its maximum value of 48.