A sector of angle $\theta$ radians and radius $4\pi\text{ cm}$ is used to form the curved surface of a right circular cone such that there is no overlapping. Prove that the volume of the cone is
$\frac{8\pi}{3}\sqrt{4\pi^2\theta^4 - \theta^6}\text{ cm}^3. \hfill [3]$ Show that the maximum volume of the cone is $\frac{p\sqrt{3}}{q}\pi^4\text{ cm}^3$ as $\theta$ varies, where $p$ and $q$ are integers to be determined. \hfill [The formula for the arc length of a sector of a circle is $s = r\theta$.]

1. Find the radius of the cone's base in terms of $\theta$.

Let $r$ be the base radius of the cone.
The arc length of the sector becomes the circumference of the cone's base.
$2\pi r = 4\pi\theta$
$r = 2\theta$

2. Express the height of the cone in terms of $\theta$.

The radius of the sector becomes the slant height of the cone, so $l = 4\pi$.
Using Pythagoras' theorem for the height $h$ of the cone:
$h = \sqrt{l^2 - r^2} = \sqrt{(4\pi)^2 - (2\theta)^2}$
$h = \sqrt{16\pi^2 - 4\theta^2} = 2\sqrt{4\pi^2 - \theta^2}$

3. Formulate and simplify the volume of the cone to prove the given expression.

$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi (2\theta)^2 \left( 2\sqrt{4\pi^2 - \theta^2} \right)$
$V = \frac{8\pi}{3}\theta^2\sqrt{4\pi^2 - \theta^2}$
Bringing $\theta^2$ inside the square root as $\theta^4$:
$V = \frac{8\pi}{3}\sqrt{\theta^4(4\pi^2 - \theta^2)}$
$V = \frac{8\pi}{3}\sqrt{4\pi^2\theta^4 - \theta^6}\text{ cm}^3 \text{ (Shown)}$

4. Differentiate the expression inside the square root to find stationary values.

Let the expression inside the square root be $u = 4\pi^2\theta^4 - \theta^6$.
To maximise $V$, we just need to maximise $u$, as $V$ is an increasing function of $u$ for $u > 0$.
$\frac{\mathrm{d}u}{\mathrm{d}\theta} = 16\pi^2\theta^3 - 6\theta^5$
For a stationary value, set $\frac{\mathrm{d}u}{\mathrm{d}\theta} = 0$:
$2\theta^3(8\pi^2 - 3\theta^2) = 0$
Since $\theta > 0$, $\theta^3 \neq 0$, so:
$8\pi^2 - 3\theta^2 = 0 \implies \theta^2 = \frac{8\pi^2}{3}$

5. Verify that this stationary value gives a maximum volume.

Using the second derivative test on $u$:
$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} = 48\pi^2\theta^2 - 30\theta^4$
Substitute $\theta^2 = \frac{8\pi^2}{3}$:
$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} = 48\pi^2\left(\frac{8\pi^2}{3}\right) - 30\left(\frac{8\pi^2}{3}\right)^2$
$\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} = 128\pi^4 - 30\left(\frac{64\pi^4}{9}\right) = 128\pi^4 - \frac{640\pi^4}{3} = -\frac{256\pi^4}{3}$
Since $\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} < 0$, $u$ (and hence $V$) is a maximum when $\theta^2 = \frac{8\pi^2}{3}$.

6. Substitute the value of $\theta^2$ back into $V$ to find $p$ and $q$.

Substitute $\theta^2 = \frac{8\pi^2}{3}$ into the equation for $V$:
$V_{\text{max}} = \frac{8\pi}{3}\sqrt{4\pi^2\left(\frac{8\pi^2}{3}\right)^2 - \left(\frac{8\pi^2}{3}\right)^3}$
$V_{\text{max}} = \frac{8\pi}{3}\sqrt{4\pi^2\left(\frac{64\pi^4}{9}\right) - \frac{512\pi^6}{27}}$
$V_{\text{max}} = \frac{8\pi}{3}\sqrt{\frac{256\pi^6}{9} - \frac{512\pi^6}{27}} = \frac{8\pi}{3}\sqrt{\frac{768\pi^6 - 512\pi^6}{27}}$
$V_{\text{max}} = \frac{8\pi}{3}\sqrt{\frac{256\pi^6}{27}} = \frac{8\pi}{3} \left( \frac{16\pi^3}{3\sqrt{3}} \right)$
$V_{\text{max}} = \frac{128\pi^4}{9\sqrt{3}} = \frac{128\sqrt{3}}{27}\pi^4$
Comparing with $\frac{p\sqrt{3}}{q}\pi^4$, we obtain:
$p = 128$ and $q = 27$.