*1 As shown in the figure, find two overlapping rectangles of height $h$ inscribed in a semicircle with radius 1. For what value of $h$ is the sum of the areas of the two rectangles maximized?
It is known that the sum of the areas is $S=2 h \sqrt{1-h^{2}}+2 h \sqrt{1-4 h^{2}}$, so we need to find the value of $h$ that maximizes $f(h)=h\left(\sqrt{1-h^{2}}+\sqrt{1-4 h^{2}}\right)$.
$$
\begin{aligned}
& h^{2}\left(\sqrt{1-h^{2}}+\sqrt{1-4 h^{2}}\right)^{2} \\
\leqslant & \left(\frac{1}{\alpha}+\frac{1}{\beta}\right) h^{2}\left[\alpha\left(1-h^{2}\right)+\beta\left(1-4 h^{2}\right)\right] \\
= & \left(\frac{1}{\alpha}+\frac{1}{\beta}\right) h^{2}\left[\alpha+\beta-(\alpha+4 \beta) h^{2}\right] \\
= & \frac{\alpha+\beta}{\alpha \beta(\alpha+4 \beta)}(\alpha+4 \beta) h^{2}\left[\alpha+\beta-(\alpha+4 \beta) h^{2}\right] \leqslant \frac{(\alpha+\beta)^{3}}{4 \alpha \beta(\alpha+4 \beta)},
\end{aligned}
$$

Here, $\alpha, \beta$ are undetermined positive numbers. The above two inequalities are the Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality, respectively. The conditions for equality, which $\alpha$ and $\beta$ must satisfy, are:
$$
\left\{\begin{array}{l}
\alpha^{2}\left(1-h^{2}\right)=\beta^{2}\left(1-4 h^{2}\right), \\
2(\alpha+4 \beta) h^{2}=\alpha+\beta,
\end{array}\right.
$$

Eliminating $h$, we get $\alpha^{2}-\beta^{2}=\left(\alpha^{2}-4 \beta^{2}\right) \frac{\alpha+\beta}{2(\alpha+4 \beta)}$. Let $\frac{\beta}{\alpha}=k$, we have $1-k^{2}=\left(1-4 k^{2}\right) \frac{1+k}{2(1+4 k)}$. Simplifying, we get $4 k^{2}-6 k-1=0, k=\frac{3+\sqrt{13}}{4}$ (negative root discarded), thus we obtain
$$
h=\sqrt{\frac{1+k}{2(1+4 k)}}=\sqrt{\frac{5-\sqrt{13}}{8}} .
$$