[The volume of a sphere of radius r is $\dfrac{4}{3}\pi r^3$, and its surface area is $4\pi r^2$.] The diagram shows a bubble tea cup of capacity $360\pi\ \text{cm}^3$. It consists of a cylindrical body of base radius $r$ cm and height $h$ cm, and a hemispheric cap. Let $S\ \text{cm}^2$ be the total surface area of the cup.

1. Formulate an equation for the total volume of the cup.

Volume of cylinder + Volume of hemisphere = $360\pi$
$\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi$

2. Express the height $h$ in terms of $r$.

$r^2 h + \frac{2}{3}r^3 = 360$
$r^2 h = 360 - \frac{2}{3}r^3$
$h = \frac{360}{r^2} - \frac{2}{3}r$

3. Formulate an expression for the total surface area $S$.

The surface area consists of the circular base, the curved surface of the cylinder, and the curved surface of the hemisphere.
$S = \pi r^2 + 2\pi r h + 2\pi r^2$
$S = 3\pi r^2 + 2\pi r h$

4. Substitute $h$ into the surface area expression and simplify to show the given result.

$S = 3\pi r^2 + 2\pi r \left( \frac{360}{r^2} - \frac{2}{3}r \right)$
$S = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2$
$S = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{720\pi}{r}$
$S = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}$
$S = \pi\left(\frac{5}{3}r^2 + \frac{720}{r}\right)$

1. Differentiate $S$ with respect to $r$.

$S = \pi\left(\frac{5}{3}r^2 + 720r^{-1}\right)$

$\frac{\text{d}S}{\text{d}r} = \pi\left(\frac{10}{3}r - 720r^{-2}\right)$

2. Set the derivative to zero to find the value of $r$ at the stationary point.

For a stationary value, $\frac{\text{d}S}{\text{d}r} = 0$.

$\pi\left(\frac{10}{3}r - \frac{720}{r^2}\right) = 0$

$\frac{10}{3}r = \frac{720}{r^2}$

$r^3 = 720 \times \frac{3}{10} = 216$

$r = \sqrt[3]{216} = 6$

3. Calculate the stationary value of $S$.

Substitute $r = 6$ into $S$:

$S = \pi\left(\frac{5}{3}(6)^2 + \frac{720}{6}\right)$

$S = \pi(60 + 120) = 180\pi$

4. Determine the nature of the stationary value using the second derivative test.

$\frac{\text{d}^2S}{\text{d}r^2} = \pi\left(\frac{10}{3} + 1440r^{-3}\right)$

When $r = 6$,
$\frac{\text{d}^2S}{\text{d}r^2} = \pi\left(\frac{10}{3} + \frac{1440}{6^3}\right) = \pi\left(\frac{10}{3} + \frac{1440}{216}\right) = 10\pi$

Since $\frac{\text{d}^2S}{\text{d}r^2} > 0$, the stationary value is a minimum.