In the isosceles triangle $ABC$, $AC = BC$, $AB = 20\text{ cm}$ and $\angle BAC = 30^\circ$. A rectangle $PQRS$ is inscribed in $ABC$ with points $P$ and $Q$ on $AB$, point $R$ on $BC$ and point $S$ on $AC$ as shown in the diagram. Taking $PS$ to be $x\text{ cm}$, show that the area of $PQRS$ may be expressed as $20x - 2\sqrt{3}x^2$. Hence, as $x$ varies, find the exact values of $PS$ and $PQ$ such that the area of $PQRS$ is a maximum, and find the corresponding area of $PQRS$.

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.

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