图中显示了一个纸镇，它由一个半径为 $r$ cm 的实心半球体，叠放在一个半径为 $2r$ cm、高为 $h$ cm 的实心圆柱体上方组成。

1. Formulate the expression for the total volume of the solid

Volume of hemisphere = $\frac{2}{3}\pi r^3$
Volume of cylinder = $\pi (2r)^2 h = 4\pi r^2 h$
Total volume = $\frac{2}{3}\pi r^3 + 4\pi r^2 h$

2. Equate the total volume to the given value and solve for $h$

$\frac{2}{3}\pi r^3 + 4\pi r^2 h = 216\pi$
Divide by $\pi$:
$\frac{2}{3}r^3 + 4r^2 h = 216$
$4r^2 h = 216 - \frac{2}{3}r^3$
$h = \frac{216}{4r^2} - \frac{2r^3}{3(4r^2)}$
$h = \frac{54}{r^2} - \frac{r}{6}$

1. Express $h$ in terms of $r$ using the volume constraint from part (a)

Total volume = Volume of hemisphere + Volume of cylinder
$\frac{2}{3}\pi r^3 + \pi(2r)^2 h = 216\pi$
$\frac{2}{3}\pi r^3 + 4\pi r^2 h = 216\pi$
$4\pi r^2 h = 216\pi - \frac{2}{3}\pi r^3$
$h = \frac{216\pi}{4\pi r^2} - \frac{\frac{2}{3}\pi r^3}{4\pi r^2}$
$h = \frac{54}{r^2} - \frac{r}{6}$

2. Formulate the expression for the total surface area of the solid

Total surface area $A$ consists of:

1. Curved surface area of hemisphere: $2\pi r^2$
2. Exposed top surface of cylinder: $\pi(2r)^2 - \pi r^2 = 3\pi r^2$
3. Curved surface area of cylinder: $2\pi(2r)h = 4\pi rh$
4. Base area of cylinder: $\pi(2r)^2 = 4\pi r^2$

$A = 2\pi r^2 + 3\pi r^2 + 4\pi rh + 4\pi r^2$
$A = 9\pi r^2 + 4\pi rh$

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

$A = 9\pi r^2 + 4\pi r\left(\frac{54}{r^2} - \frac{r}{6}\right)$
$A = 9\pi r^2 + \frac{216\pi}{r} - \frac{4\pi r^2}{6}$
$A = 9\pi r^2 - \frac{2}{3}\pi r^2 + \frac{216\pi}{r}$
$A = \frac{25\pi r^2}{3} + \frac{216\pi}{r}$ (Shown)

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

$A = \frac{25\pi r^2}{3} + 216\pi r^{-1}$

$\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{50\pi r}{3} - 216\pi r^{-2}$

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

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$:
$\frac{50\pi r}{3} - \frac{216\pi}{r^2} = 0$
$\frac{50\pi r}{3} = \frac{216\pi}{r^2}$
$50r^3 = 648$
$r^3 = 12.96$
$r = \sqrt[3]{12.96} \approx 2.3489 \approx 2.35$ (to 3 s.f.)

3. Calculate the corresponding stationary value of $A$.

Substitute $r = \sqrt[3]{12.96}$ into $A$:
$A = \frac{25\pi (12.96)^{2/3}}{3} + \frac{216\pi}{(12.96)^{1/3}}$
$A \approx 433.43 \approx 433$ (to 3 s.f.)

4. Use the second derivative test to determine the nature of the stationary value.

$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{50\pi}{3} + 432\pi r^{-3}$
Since $r > 0$, $\frac{\mathrm{d}^2A}{\mathrm{d}r^2} > 0$.
Therefore, the stationary value is a minimum.