Example 5. Find the maximum area of a trapezoid inscribed in an ellipse with the major axis as one base.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $P(a \cos \theta, b \sin \theta)$.
By symmetry, $P^{\prime}(-a \cos \theta, b \sin \theta)$.
$$
\begin{array}{l}
\therefore \sqrt{3 \cos ^{3} \frac{\theta}{2} \cdot \sin ^{2} \frac{\theta}{2}} \leqslant\left(\frac{3}{4}\right)^{2} \text {. } \\
\end{array}
$$

Therefore, the area of the trapezoid $S$
$$
\begin{array}{l}
=\frac{4 \sqrt{3}}{3} a b \sqrt{3 \cos ^{3} \frac{\theta}{2} \sin ^{2} \frac{\theta}{2}} \\
\leqslant-\frac{4 \sqrt{3}}{3} \cdot a b\left(\frac{3}{4}\right)^{2}=\frac{3 \sqrt{3} a b}{4} .
\end{array}
$$

That is, the maximum area of the trapezoid inscribed in the ellipse with the major axis as one base is $\frac{3 \sqrt{3} a b}{4}$.

This is a comprehensive problem involving analytic geometry, trigonometry, and algebra, using coordinates to find the area of the trapezoid, applying trigonometric formulas for transformation, and using basic inequalities to find the extremum.
(Author's address: Tianjin Experimental High School)