2. As shown in Figure 1, given $A(-3,0)$, $B(0,-4)$, and $P$ is any point on the hyperbola $y=\frac{12}{x} (x > 0)$. A perpendicular line from point $P$ to the x-axis meets at point $C$, and a perpendicular line from point $P$ to the y-axis meets at point $D$. Then the minimum value of the area $S$ of quadrilateral $A B C D$ is ( ).
(A) 22
(B) 23
(C) 24
(D) 26
2. C.

Let $P\left(x, \frac{12}{x}\right)$. Then $C(x, 0)$ and $D\left(0, \frac{12}{x}\right)$.
It is easy to see that $C A=x+3, D B=\frac{12}{x}+4$, so
$$
S=\frac{1}{2} C A \cdot D B=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \text {. }
$$

Simplifying, we get $S=2\left(x+\frac{9}{x}\right)+12$.
Since $x>0, \frac{9}{x}>0$, we have
$$
x+\frac{9}{x}=\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)^{2}+6 \geqslant 6 \text {. }
$$

Equality holds only when $x=\frac{9}{x}$, i.e., $x=3$. Therefore, $S \geqslant 2 \times 6+12=24$.