Example 4 For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the endpoints of the major axis are $A$ and $B$, and $C, D$ are two points on the ellipse on the same side of $A B$, with $C D / / A B$. Find the maximum value of the area of trapezoid $A B C D$.
Solution: From $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, 
we can set $x=a \cos \theta, y=$ $b \sin \theta, \theta$ as the parameter. As shown in Figure 2, let $C(a \cos \theta, b \sin \theta)$. According to the problem, we have
$$
\begin{array}{l}
=a b(1+\cos \theta) \sqrt{1-\cos ^{2} \theta} \\
=a b \sqrt{(1+\cos \theta)^{3}}(1-\cos \theta) \\
=\frac{a b}{\sqrt{3}} \cdot \sqrt{(1+\cos \theta)^{3}(3-3 \cos \theta)} \\
\leqslant \frac{a b}{\sqrt{3}}\left(\frac{3+3 \cos \theta+3-3 \cos \theta}{4}\right)^{2} \\
=\frac{3 \sqrt{3}}{4} a b \text {. } \\
\end{array}
$$

Equality holds if and only if $1+\cos \theta=3-3 \cos \theta$, i.e., $\cos \theta=\frac{1}{2}$.
Therefore, the maximum value of the trapezoid area is $\frac{3 \sqrt{3}}{4} a b$.