$ABCD$ 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 length of $OB$ in terms of $x$.

Since $O$ is the midpoint of $AB$, $OB = \frac{x}{2}$.

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

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

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

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

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

$A = \frac{1}{2}x(256 - x^2)^{\frac{1}{2}}$

$\frac{\text{d}A}{\text{d}x} = \frac{1}{2}(256 - x^2)^{\frac{1}{2}} + \frac{1}{2}x \left[ \frac{1}{2}(256 - x^2)^{-\frac{1}{2}}(-2x) \right]$
$\frac{\text{d}A}{\text{d}x} = \frac{1}{2}(256 - x^2)^{\frac{1}{2}} - \frac{x^2}{2(256 - x^2)^{\frac{1}{2}}}$

2. Set the derivative to zero to find the stationary value.

For a stationary value, $\frac{\text{d}A}{\text{d}x} = 0$:
$\frac{1}{2}(256 - x^2)^{\frac{1}{2}} - \frac{x^2}{2(256 - x^2)^{\frac{1}{2}}} = 0$
$\frac{1}{2}(256 - x^2)^{\frac{1}{2}} = \frac{x^2}{2(256 - x^2)^{\frac{1}{2}}}$

3. Solve the equation for $x$.

Multiply both sides by $2(256 - x^2)^{\frac{1}{2}}$:
$256 - x^2 = x^2$
$2x^2 = 256$
$x^2 = 128$
$x = \sqrt{128} = 8\sqrt{2}$ (since $x > 0$ as it is a length)
$x \approx 11.3$

1. Evaluate the sign of the first derivative slightly before and after the stationary point

From part (b), the stationary value occurs at $x = 8\sqrt{2} \approx 11.31$.
Let us test values of $x$ slightly less than and slightly greater than $8\sqrt{2}$.
$\frac{\text{d}A}{\text{d}x} = \frac{256 - 2x^2}{2\sqrt{256 - x^2}}$
When $x = 11$, $\frac{\text{d}A}{\text{d}x} = \frac{256 - 2(11)^2}{2\sqrt{256 - 11^2}} = \frac{14}{2\sqrt{135}} > 0$
When $x = 11.5$, $\frac{\text{d}A}{\text{d}x} = \frac{256 - 2(11.5)^2}{2\sqrt{256 - 11.5^2}} = \frac{-8.5}{2\sqrt{123.75}} < 0$

2. Conclude the nature of the stationary point

Since $\frac{\text{d}A}{\text{d}x}$ changes from positive to negative as $x$ increases through $8\sqrt{2}$, the stationary value is a maximum.