一个高为 $h$、底面半径为 $r$ 的圆锥恰好内接于一个半径为 $5\text{ cm}$ 的球体中，如下图所示。 $\left(\text{圆锥的体积} = \frac{1}{3}\pi r^2 h\right)$

1. Identify the right-angled triangle relating $r$, $h$, and the sphere's radius.

Let the center of the sphere be $O$. The distance from $O$ to the base of the cone is $h - 5$. The radius of the base is $r$, and the radius of the sphere is $5$. These form a right-angled triangle.

2. Apply Pythagoras' theorem to express $r$ in terms of $h$.

$r^2 + (h - 5)^2 = 5^2$
$r^2 + h^2 - 10h + 25 = 25$
$r^2 = 10h - h^2$
$r = \sqrt{10h - h^2}$

1. Recall or derive the expression for $r^2$ in terms of $h$ from part (i)

From part (i), using Pythagoras' theorem on the triangle formed by the sphere's centre, the cone's base centre, and the edge of the base:
$r^2 + (h - 5)^2 = 5^2$
$r^2 + h^2 - 10h + 25 = 25$
$r^2 = 10h - h^2$

2. Substitute $r^2$ into the volume formula to show the given expression

$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi (10h - h^2) h$
$V = \frac{\pi h^2 (10 - h)}{3}$ (Shown)

1. Differentiate $V$ with respect to $h$

$V = \frac{\pi}{3}(10h^2 - h^3)$

$\frac{dV}{dh} = \frac{\pi}{3}(20h - 3h^2)$

2. Set the derivative to zero to find stationary points

For stationary $V$, $\frac{dV}{dh} = 0$

$\frac{\pi}{3}(20h - 3h^2) = 0$

$h(20 - 3h) = 0$

3. Solve for $h$ and reject the invalid solution

$h = 0$ or $h = \frac{20}{3}$

Since $h > 0$ for a cone, $h = \frac{20}{3}$

1. Substitute the stationary value of $h$ into the volume equation

From part (iii), the stationary value is $h = \frac{20}{3}$.
$V = \frac{\pi \left(\frac{20}{3}\right)^2 \left(10 - \frac{20}{3}\right)}{3}$
$V = \frac{\pi \left(\frac{400}{9}\right) \left(\frac{10}{3}\right)}{3}$
$V = \frac{4000\pi}{81} \approx 155.14 \text{ cm}^3$

2. Find the second derivative of $V$ with respect to $h$

From part (iii), $\frac{\text{d}V}{\text{d}h} = \frac{20\pi}{3}h - \pi h^2$.
$\frac{\text{d}^2V}{\text{d}h^2} = \frac{20\pi}{3} - 2\pi h$

3. Evaluate the second derivative at $h = \frac{20}{3}$ to determine the nature of the stationary point

When $h = \frac{20}{3}$,
$\frac{\text{d}^2V}{\text{d}h^2} = \frac{20\pi}{3} - 2\pi \left(\frac{20}{3}\right) = -\frac{20\pi}{3}$
Since $\frac{\text{d}^2V}{\text{d}h^2} < 0$, the volume is a maximum.