20. (12 points) Let $f(x)$

be a function defined on $\mathbf{R}$ with a period of 2, and it is an even function. On the interval $[2,3]$, $f(x)=-2(x-3)^{2}+4$. Rectangle $A B C D$ has two vertices $A$ and $B$ on the $x$-axis, and $C$ and $D$ on the graph of $y=f(x)(0 \leqslant x \leqslant 2)$. Find the maximum area of rectangle $A B C D$.
20. When $0 \leqslant x \leqslant 1$, we have
$$
f(x)=f(x+2)=-2(x-1)^{2}+4 \text {; }
$$

When $-1 \leqslant x \leqslant 0$, we have
$$
f(x)=f(-x)=-2(-x-1)^{2}+4 \text {; }
$$

When $1 \leqslant x \leqslant 2$, we have
$$
\begin{array}{l}
f(x)=f(x-2)=-2[-(x-2)-1]^{2}+4 \\
=-2(x-1)^{2}+4 .
\end{array}
$$

Let $D(x, t) 、 C(2-x, t)$.
Then $t=-2(x-1)^{2}+4$. It is easy to see that
$$
\begin{array}{l}
S_{\text {short } 1 B C D}=|A B| \cdot|B C| \\
=(2-2 x) t=\sqrt{(8-2 t) t \cdot t} \\
\leqslant \sqrt{\left[\frac{(8-2 t)+t+t}{3}\right]^{3}}=\frac{16 \sqrt{6}}{9} .
\end{array}
$$

When and only when $t=\frac{8}{3}$, i.e., $x=1-\frac{\sqrt{6}}{3}$, the area of rectangle $A B C D$ reaches its maximum value $\frac{16 \sqrt{6}}{9}$.