某公司制造一个如下图所示的玻璃密封容器。 该密封容器的厚度忽略不计，由两个部分组成。底部是一个底面半径为 $r\text{ cm}$、高为 $h\text{ cm}$ 的圆柱体。顶部是一个半径为 $r\text{ cm}$ 的半球。公司要求容器的体积固定为 $108\pi\text{ cm}^3$。 容器的外表面积记为 $A\text{ cm}^2$。为了降低生产成本，公司希望 $A$ 的值尽可能小。 [半径为 $r$ 的球体体积公式为 $\frac{4}{3}\pi r^3$ 并其表面积公式为 $4\pi r^2$。]

1. Write down the equation for the total volume of the container and equate it to $108\pi$.

$\text{Volume of container} = \text{Volume of cylinder} + \text{Volume of hemisphere}$
$\pi r^2 h + \frac{2}{3}\pi r^3 = 108\pi$

2. Rearrange the volume equation to make $h$ the subject.

Divide both sides by $\pi$:
$r^2 h + \frac{2}{3}r^3 = 108$
$r^2 h = 108 - \frac{2}{3}r^3$
$h = \frac{108}{r^2} - \frac{2r}{3}$

3. Formulate the external surface area, substitute the expression for $h$, and simplify.

$\text{External surface area } A = \text{Base area} + \text{Cylinder curved area} + \text{Hemisphere curved area}$
$A = \pi r^2 + 2\pi r h + 2\pi r^2$
$A = 3\pi r^2 + 2\pi r h$
Substitute $h = \frac{108}{r^2} - \frac{2r}{3}$:
$A = 3\pi r^2 + 2\pi r \left( \frac{108}{r^2} - \frac{2r}{3} \right)$
$A = 3\pi r^2 + \frac{216\pi}{r} - \frac{4\pi r^2}{3}$
$A = \frac{5\pi r^2}{3} + \frac{216\pi}{r}$
$A = \left( \frac{5r^2}{3} + \frac{216}{r} \right)\pi \quad \text{(shown)}$

1. Find the first derivative of $A$ with respect to $r$.

$A = \left(\frac{5}{3}r^2 + 216r^{-1}\right)\pi$

$\frac{\mathrm{d}A}{\mathrm{d}r} = \pi\left(\frac{10}{3}r - 216r^{-2}\right) = \frac{10\pi}{3}r - \frac{216\pi}{r^2}$

2. Equate the first derivative to zero to find the exact value of $r$.

For minimum $A$, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$
$\frac{10\pi}{3}r - \frac{216\pi}{r^2} = 0$
$\frac{10}{3}r = \frac{216}{r^2}$
$10r^3 = 648$
$r^3 = 64.8$
$r = \sqrt[3]{64.8} \text{ (or } \sqrt[3]{\frac{324}{5}} \text{)}$

3. Find the second derivative of $A$ to determine the nature of the stationary point.

$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \pi\left(\frac{10}{3} - (-2)(216)r^{-3}\right)$
$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10\pi}{3} + \frac{432\pi}{r^3}$

4. Substitute the value of $r$ into the second derivative to confirm it is a minimum.

Substitute $r^3 = 64.8$ into the second derivative:
$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10\pi}{3} + \frac{432\pi}{64.8}$
$\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = \frac{10\pi}{3} + \frac{20\pi}{3} = 10\pi$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}r^2} > 0$, the value of $A$ is a minimum.

1. Establish the valid domain for $r$ using the physical constraint that the cylinder's height cannot be negative.

From the volume constraint $108\pi = \pi r^2 h + \frac{2}{3}\pi r^3$, we can make the height $h$ the subject:
$h = \frac{108}{r^2} - \frac{2}{3}r$
Since the physical container must have a non-negative cylindrical height ($h \ge 0$):
$\frac{108}{r^2} - \frac{2}{3}r \ge 0$
$r^3 \le 162$
$r \le \sqrt[3]{162}$
As $r > 0$ for a closed container to exist, the valid domain is $0 < r \le \sqrt[3]{162}$.

2. Calculate the exact coordinates of the end point when $r$ is at its maximum.

The graph terminates when $r^3 = 162$ (or $r = \sqrt[3]{162} = 3\sqrt[3]{6}$).
Substitute this into $A = \pi \left(\frac{5r^2}{3} + \frac{216}{r}\right)$.
Note that when $r^3 = 162$, we can rewrite the second term as:
$\frac{216}{r} = \frac{216r^2}{r^3} = \frac{216r^2}{162} = \frac{4}{3}r^2$
Thus, the area simplifies nicely:
$A = \pi \left(\frac{5}{3}r^2 + \frac{4}{3}r^2\right) = 3\pi r^2$
Substitute $r = 3\sqrt[3]{6}$:
$A = 3\pi (3\sqrt[3]{6})^2 = 3\pi (9 \cdot 6^{2/3}) = 27\pi\sqrt[3]{36}$
The coordinates of the end point are $(\sqrt[3]{162}, 27\pi\sqrt[3]{36})$ or approximately $(5.45, 280)$.

3. Determine the features of the sketch including the asymptote, the minimum point, and the end point.

To sketch the graph accurately:

• Vertical Asymptote: As $r \to 0$, $\frac{216}{r} \to \infty$, so $A \to \infty$. The $A$-axis is a vertical asymptote.
• Minimum Point: From part (b), $\frac{dA}{dr} = 0$ gives a minimum point at $r = \sqrt[3]{64.8} \approx 4.02$, with $A \approx 253.1$.
• End Point: The graph ends definitively at the solid point $(\sqrt[3]{162}, 27\pi\sqrt[3]{36})$.
  The curve must show a smooth, convex (U-shaped) profile dropping from infinity to the minimum, then curving slightly upwards to terminate abruptly at the boundary.

1. Determine the position of the liquid level at $t = 14\text{ s}$.

The volume of the liquid leaked after $14\text{ s}$ is:
$V_{\text{leaked}} = 14 \times 5 = 70 \text{ cm}^3$

The volume of the hemispherical top is:
$V_{\text{hemisphere}} = \frac{2}{3}\pi (3)^3 = 18\pi \approx 56.55 \text{ cm}^3$

Since $V_{\text{leaked}} > V_{\text{hemisphere}}$, the hemisphere has completely emptied. The liquid level is therefore entirely within the cylindrical part of the container.

2. Express the volume of the liquid in the cylinder in terms of its height.

Let $h$ be the height of the liquid from the bottom. Since the liquid is in the cylindrical section, its volume $V$ is:
$V = \pi r^2 h = \pi (3)^2 h = 9\pi h$

3. Differentiate the volume with respect to time to find the rate of decrease of the height.

Differentiating both sides with respect to $t$:
$\frac{\mathrm{d}V}{\mathrm{d}t} = 9\pi \frac{\mathrm{d}h}{\mathrm{d}t}$

Given that the liquid leaks at a constant rate of $5 \text{ cm}^3/\text{s}$, we have $\frac{\mathrm{d}V}{\mathrm{d}t} = -5$:
$-5 = 9\pi \frac{\mathrm{d}h}{\mathrm{d}t}$
$\frac{\mathrm{d}h}{\mathrm{d}t} = -\frac{5}{9\pi}$

The rate of decrease of the height is $\frac{5}{9\pi} \text{ cm/s}$ (or $0.177 \text{ cm/s}$ to 3 s.f.).