在等腰三角形 $ABC$ 中，$AC = BC$，$AB = 20\text{ cm}$，且 $\angle BAC = 30^\circ$。一个矩形 $PQRS$ 内接于 $ABC$，其中顶点 $P$ 和 $Q$ 在 $AB$ 上，顶点 $R$ 在 $BC$ 上，顶点 $S$ 在 $AC$ 上，如图所示。 设 $PS$ 为 $x\text{ cm}$，证明 $PQRS$ 的面积可以表示为 $20x - 2\sqrt{3}x^2$。 由此，当 $x$ 变化时，求使得 $PQRS$ 面积最大时 $PS$ 和 $PQ$ 的精确值，并求出 $PQRS$ 的对应面积。

1. Relate the base lengths $AP$ and $QB$ to $x$.

In the right-angled triangle $APS$,
$\tan 30^\circ = \frac{PS}{AP}$
$AP = \frac{x}{\tan 30^\circ} = x \div \frac{1}{\sqrt{3}} = \sqrt{3}x$
Since $AC = BC$, triangle $ABC$ is isosceles, which means $\angle ABC = \angle BAC = 30^\circ$.
In the right-angled triangle $BQR$, $QR = PS = x$, so
$\tan 30^\circ = \frac{QR}{QB} \implies QB = \sqrt{3}x$

2. Express the length of $PQ$ and show the given area of the rectangle.

The length of the rectangle is $PQ$.
$PQ = AB - AP - QB = 20 - \sqrt{3}x - \sqrt{3}x = 20 - 2\sqrt{3}x$
The area of the rectangle $PQRS$ is given by $A = PQ \times PS$.
$A = (20 - 2\sqrt{3}x)x = 20x - 2\sqrt{3}x^2 \text{ (shown)}$

3. Differentiate the area with respect to $x$ to find the stationary point.

Let $A = 20x - 2\sqrt{3}x^2$.
Differentiating with respect to $x$:
$\frac{\mathrm{d}A}{\mathrm{d}x} = 20 - 4\sqrt{3}x$
For a stationary point (maximum/minimum area), set $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$:
$20 - 4\sqrt{3}x = 0$
$4\sqrt{3}x = 20 \implies x = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$

4. Verify the nature of the stationary point and find exact values of $PS$ and $PQ$.

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -4\sqrt{3}$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} < 0$, the area $A$ is indeed a maximum when $x = \frac{5\sqrt{3}}{3}$.
$PS = x = \frac{5\sqrt{3}}{3}\text{ cm}$
$PQ = 20 - 2\sqrt{3}\left(\frac{5\sqrt{3}}{3}\right) = 20 - 10 = 10\text{ cm}$

5. Calculate the corresponding maximum area.

$\text{Maximum Area} = PS \times PQ$
$= \frac{5\sqrt{3}}{3} \times 10 = \frac{50\sqrt{3}}{3}\text{ cm}^2$