$ABCD$ 是一个位于半径为 8 cm、圆心为 $O$ 的半圆内的矩形。已知 $AB = x$ cm 且 $BC = y$ cm。

1. Express the distance from the centre $O$ to $B$ in terms of $x$.

Since $ABCD$ is a rectangle inscribed in a semicircle with centre $O$ on $AB$, $O$ is the midpoint of $AB$.
$OB = \frac{x}{2}$

2. Use Pythagoras' theorem to express $y$ in terms of $x$.

In $\triangle OBC$, $\angle OBC = 90^\circ$.
$OB^2 + BC^2 = OC^2$
$\left(\frac{x}{2}\right)^2 + y^2 = 8^2$
$y^2 = 64 - \frac{x^2}{4}$
$y = \sqrt{64 - \frac{x^2}{4}} = \sqrt{\frac{256 - x^2}{4}} = \frac{1}{2}\sqrt{256 - x^2}$

3. Formulate the area of the rectangle $A$ in terms of $x$.

$A = AB \times BC$
$A = x \times y$
$A = x \left( \frac{1}{2}\sqrt{256 - x^2} \right)$
$A = \frac{x}{2}\sqrt{256 - x^2}$ (Shown)

1. Differentiate $A$ with respect to $x$ using the product rule and chain rule.

$A = \frac{x}{2}(256 - x^2)^{\frac{1}{2}}$
$\frac{\text{d}A}{\text{d}x} = \frac{1}{2}(256 - x^2)^{\frac{1}{2}} + \frac{x}{2} \cdot \frac{1}{2}(256 - x^2)^{-\frac{1}{2}}(-2x)$

2. Simplify the derivative.

$\frac{\text{d}A}{\text{d}x} = \frac{1}{2}\sqrt{256 - x^2} - \frac{x^2}{2\sqrt{256 - x^2}}$
$\frac{\text{d}A}{\text{d}x} = \frac{256 - x^2 - x^2}{2\sqrt{256 - x^2}} = \frac{256 - 2x^2}{2\sqrt{256 - x^2}} = \frac{128 - x^2}{\sqrt{256 - x^2}}$

3. Set the derivative to zero to find the stationary value of $x$.

For a stationary value, $\frac{\text{d}A}{\text{d}x} = 0$.
$\frac{128 - x^2}{\sqrt{256 - x^2}} = 0$
$128 - x^2 = 0$
$x^2 = 128$
$x = \sqrt{128} = 8\sqrt{2}$ (since $x > 0$ as it is a length)

1. Recall the expression for $\frac{\text{d}A}{\text{d}x}$ and the stationary value of $x$ from part (b)

From part (b), $\frac{\text{d}A}{\text{d}x} = \frac{128 - x^2}{\sqrt{256 - x^2}}$.
The stationary value occurs at $x = \sqrt{128} \approx 11.31$.

2. Test the sign of $\frac{\text{d}A}{\text{d}x}$ slightly before and after the stationary point

Let $x = 11$ (slightly less than $\sqrt{128}$):
$\frac{\text{d}A}{\text{d}x} = \frac{128 - 11^2}{\sqrt{256 - 11^2}} = \frac{128 - 121}{\sqrt{135}} > 0$

Let $x = 11.5$ (slightly more than $\sqrt{128}$):
$\frac{\text{d}A}{\text{d}x} = \frac{128 - 11.5^2}{\sqrt{256 - 11.5^2}} = \frac{128 - 132.25}{\sqrt{123.75}} < 0$

3. Conclude the nature of the stationary point based on the sign change

Since $\frac{\text{d}A}{\text{d}x}$ changes from positive to negative as $x$ increases through $\sqrt{128}$, the stationary value of $A$ is a maximum.