Example 1. Let $a x+b y=c\left(a, b, c \in R^{+}, \boldsymbol{x}\right.$ , $\left.y \in \overline{R^{-}}\right)$, find the extremum of $f(x, y)=m \sqrt{x}+n \sqrt{y}(m$, $n>0)$.
Consider the points $A(\sqrt{a x},-\sqrt{b y})$, $B\left(\frac{n}{\sqrt{b}}, \frac{m}{\sqrt{a}}\right)$, and $\angle A O B=\theta$, then
$$
\begin{array}{l}
f(x, y)=O A \cdot O B \sin \theta \\
=\sqrt{(0-\sqrt{a x})^{2}+(0+\sqrt{b y})^{2}} \\
\cdot \sqrt{\left(0-\frac{n}{\sqrt{b}}\right)^{2}+\left(0-\frac{m}{\sqrt{a}}\right)^{2}} \cdot \sin \theta \\
=\sqrt{\frac{a n^{2}+b m^{2}}{a b}} \sqrt{c} \sin \theta .
\end{array}
$$

Let $k=\min \left(\frac{n}{\sqrt{b}},-\frac{m}{\sqrt{a}}\right)$, then
$$
\begin{array}{l}
k / \sqrt{\frac{a n^{2}+b m^{2}}{a b}} \leqslant \sin \theta \leqslant 1 . \\
\therefore f_{\text{max}}(x, y)=\sqrt{\frac{c}{a b}\left(a n^{2}+b m^{2}\right)}, \\
\quad f_{\text{min}}(x, y)=\sqrt{c} k .
\end{array}
$$