图中所示为一个玻璃试管，其半径为 $r\text{ cm}$，高度为 $(h+r)\text{ cm}$，底部为一个半径为 $r\text{ cm}$ 的半球。试管的表面积为 $12\pi$。

1. Formulate the expression for the total surface area of the test tube

The test tube consists of an open cylinder and a hemispherical base.
Curved surface area of cylinder = $2\pi rh$
Curved surface area of hemisphere = $\frac{1}{2}(4\pi r^2) = 2\pi r^2$
Total surface area = $2\pi rh + 2\pi r^2$

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

$2\pi rh + 2\pi r^2 = 12\pi$
Divide by $2\pi$:
$rh + r^2 = 6$
$rh = 6 - r^2$
$h = \frac{6-r^2}{r}$

1. Write down the formula for the total volume of the test tube

The volume $V$ consists of the volume of the cylinder and the volume of the hemisphere.
$V = \pi r^2 h + \frac{2}{3}\pi r^3$

2. Substitute the expression for h from part (i) into the volume formula and simplify

Substitute $h = \frac{6-r^2}{r}$:
$V = \pi r^2 \left( \frac{6-r^2}{r} \right) + \frac{2}{3}\pi r^3$
$V = \pi r (6 - r^2) + \frac{2}{3}\pi r^3$
$V = 6\pi r - \pi r^3 + \frac{2}{3}\pi r^3$
$V = 6\pi r - \frac{1}{3}\pi r^3$ (Shown)

1. Differentiate $V$ with respect to $r$ to find $\frac{dV}{dr}$.

$V = 6\pi r - \frac{1}{3}\pi r^3$
$\frac{dV}{dr} = 6\pi - \pi r^2$

2. Set $\frac{dV}{dr} = 0$ to find the stationary value of $r$.

For stationary value, $\frac{dV}{dr} = 0$
$6\pi - \pi r^2 = 0$
$\pi r^2 = 6\pi$
$r^2 = 6$
$r = \sqrt{6}$ (since $r > 0$)

3. Calculate the stationary value of $V$.

Substitute $r = \sqrt{6}$ into $V$:
$V = 6\pi(\sqrt{6}) - \frac{1}{3}\pi(\sqrt{6})^3$
$V = 6\pi\sqrt{6} - \frac{1}{3}\pi(6\sqrt{6})$
$V = 6\pi\sqrt{6} - 2\pi\sqrt{6}$
$V = 4\pi\sqrt{6} \approx 30.8\text{ cm}^3$ (or exact $4\pi\sqrt{6}$)

4. Find the second derivative $\frac{d^2V}{dr^2}$ to determine the nature of the stationary point.

$\frac{d^2V}{dr^2} = -2\pi r$

5. Apply the second derivative test.

When $r = \sqrt{6}$, $\frac{d^2V}{dr^2} = -2\pi(\sqrt{6}) < 0$.
Since $\frac{d^2V}{dr^2} < 0$, the volume is a maximum.