4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is
4. 24.

Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$

The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24.