一个胶囊由一个圆柱体和在其一端的半球体组成。
圆柱体的高为 $h\text{ cm}$，半径为 $r\text{ cm}$。
半球体的半径也为 $r\text{ cm}$。 （球体体积 = $\frac{4}{3}\pi r^3$，球体表面积 = $4\pi r^2$，
圆柱体体积 = $\pi r^2 h$，圆柱体侧面积 = $2\pi r h$） 已知该胶囊的体积为 $100\pi\text{ cm}^3$。

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

Volume of capsule = Volume of cylinder + Volume of hemisphere
$V = \pi r^2 h + \frac{1}{2} \left( \frac{4}{3}\pi r^3 \right)$
$V = \pi r^2 h + \frac{2}{3}\pi r^3$

2. Equate the volume to the given value and make $h$ the subject

Given $V = 100\pi$:
$\pi r^2 h + \frac{2}{3}\pi r^3 = 100\pi$
Divide both sides by $\pi$:
$r^2 h + \frac{2}{3}r^3 = 100$
$r^2 h = 100 - \frac{2}{3}r^3$
$h = \frac{100}{r^2} - \frac{2}{3}r$ (Shown)

1. Formulate the expression for the total surface area of the capsule

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

2. Substitute the expression for $h$ from part (a) into the area equation

Substitute $h = \frac{100}{r^2} - \frac{2}{3}r$:
$A = 3\pi r^2 + 2\pi r\left(\frac{100}{r^2} - \frac{2}{3}r\right)$

3. Expand and simplify to show the required result

$A = 3\pi r^2 + \frac{200\pi}{r} - \frac{4\pi}{3}r^2$
$A = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{200\pi}{r}$
$A = \frac{5\pi}{3}r^2 + \frac{200\pi}{r}$ (Shown)

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

$A = \frac{5\pi}{3}r^2 + 200\pi r^{-1}$
$\frac{\mathrm{d}A}{\mathrm{d}r} = \frac{10\pi}{3}r - 200\pi r^{-2}$

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

For stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$
$\frac{10\pi}{3}r - \frac{200\pi}{r^2} = 0$
$\frac{10\pi}{3}r = \frac{200\pi}{r^2}$

3. Solve for $r$

$r^3 = \frac{200\pi \times 3}{10\pi}$
$r^3 = 60$
$r = \sqrt[3]{60} \approx 3.9148 \approx 3.91$ (to 3 s.f.)

4. Find the second derivative to determine the nature of the stationary point

$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10\pi}{3} - (-2)(200\pi)r^{-3}$
$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10\pi}{3} + \frac{400\pi}{r^3}$

5. Evaluate the second derivative at the stationary point to conclude

When $r = \sqrt[3]{60}$, $r > 0$, so $\frac{400\pi}{r^3} > 0$.
Thus, $\frac{\mathrm{d}^2A}{\mathrm{d}r^2} > 0$.
Since the second derivative is positive, the value of $r$ gives a minimum value of $A$.