[球体的表面积和体积分别由 $4\pi r^2$ 和 $\dfrac{4}{3}\pi r^3$ 给出。] 一位预算固定为 $k 的水族爱好者想要建造一个金鱼避难所（鱼缸）。他设计的鱼缸包括一个圆形顶盖、一个长度为 $L$ cm 的圆柱体和一个半径为 $r$ cm 的半球形底座，它们紧密地贴合在一起。用于顶盖、圆柱面和半球形底座的材料每 cm$^2$ 的成本分别为 $20、$10 和 $30。假设制作鱼缸的材料厚度可以忽略不计。

1. Express the total material cost in terms of $r$ and $L$ and equate it to $k$.

Calculate the cost of each part of the tank using the given cost rates and surface area formulas:
Cost of circular cover $= 20 \times (\pi r^2) = 20\pi r^2$
Cost of cylindrical surface $= 10 \times (2\pi r L) = 20\pi r L$
Cost of hemispherical base $= 30 \times \left(\frac{1}{2} \times 4\pi r^2\right) = 60\pi r^2$

Total cost of material $= 20\pi r^2 + 20\pi r L + 60\pi r^2 = 80\pi r^2 + 20\pi r L$

Given the fixed budget is $\char36{}k$:
$80\pi r^2 + 20\pi r L = k$

2. Make $L$ the subject of the equation.

$20\pi r L = k - 80\pi r^2$
$L = \frac{k - 80\pi r^2}{20\pi r}$
$L = \frac{k}{20\pi r} - \frac{80\pi r^2}{20\pi r}$
$L = \frac{k}{20\pi r} - 4r$

3. Set up the equation for the total volume and substitute $L$ to show the target expression.

Total volume $V$ is the sum of the volume of the cylinder and the hemisphere:
$V = \pi r^2 L + \frac{2}{3}\pi r^3$

Substitute $L = \frac{k}{20\pi r} - 4r$ into the volume equation:
$V = \pi r^2 \left( \frac{k}{20\pi r} - 4r \right) + \frac{2}{3}\pi r^3$
$V = \frac{k}{20}r - 4\pi r^3 + \frac{2}{3}\pi r^3$
$V = \frac{k}{20}r - \frac{10}{3}\pi r^3 \quad \text{(Shown)}$

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

$\frac{\mathrm{d}V}{\mathrm{d}r} = \frac{k}{20} - \frac{10}{3}\pi (3r^2) = \frac{k}{20} - 10\pi r^2$

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

$\frac{k}{20} - 10\pi r^2 = 0$
$10\pi r^2 = \frac{k}{20}$
$r^2 = \frac{k}{200\pi}$

3. Use the second derivative to verify that this value of $r$ gives a maximum volume.

$\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -20\pi r$
Since the radius $r > 0$, we have $\frac{\mathrm{d}^2V}{\mathrm{d}r^2} < 0$.
Therefore, $V$ is at a maximum.

4. Formulate the cost of the hemispherical base and evaluate it using the stationary value of $r^2$.

The surface area of the hemispherical base is $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$.

$\text{Cost} = \$30 \times 2\pi r^2 = 60\pi r^2$
Substitute $r^2 = \frac{k}{200\pi}$ into the cost equation:
$\text{Cost} = 60\pi \left(\frac{k}{200\pi}\right) = \frac{60}{200}k = \frac{3}{10}k$

1. Formulate a relationship between the radius of the water surface $x$ and the depth $h$.

Let $O$ be the centre of the hemispherical base. Since the radius of the cylinder is $15\text{ cm}$, the radius of the hemisphere is also $15\text{ cm}$.

The vertical distance from $O$ to the water surface is $15 - h$.
By Pythagoras' theorem on the vertical cross-section:
$x^2 + (15 - h)^2 = 15^2$
$x^2 + 225 - 30h + h^2 = 225$
$x^2 = 30h - h^2$

2. Express the volume of the water $W$ solely in terms of $h$.

Substitute $x^2 = 30h - h^2$ into the given volume formula $W = \frac{\pi h(3x^2 + h^2)}{6}$:
$W = \frac{\pi h(3(30h - h^2) + h^2)}{6}$
$W = \frac{\pi h(90h - 3h^2 + h^2)}{6}$
$W = \frac{\pi h(90h - 2h^2)}{6}$
$W = \pi\left(15h^2 - \frac{1}{3}h^3\right)$

3. Differentiate $W$ with respect to $h$ and evaluate at $h=3$.

$\frac{dW}{dh} = \pi\left(30h - h^2\right)$
When $h = 3$:
$\frac{dW}{dh} = \pi(30(3) - 3^2) = \pi(90 - 9) = 81\pi$

4. Use the chain rule to find the rate of change of depth $\frac{dh}{dt}$.

Since water is leaking, the rate of change of volume is $\frac{dW}{dt} = -20\text{ cm}^3/\text{min}$.
Using the chain rule:
$\frac{dW}{dt} = \frac{dW}{dh} \times \frac{dh}{dt}$
$-20 = 81\pi \times \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{20}{81\pi} \approx -0.0786\text{ cm/min}$

1. Formulate the relationship between $x$ and $h$ using Pythagoras' theorem on the hemispherical base.

From the cross-section of the hemisphere (radius $r = 15$ cm), the distance from the centre to the water surface is $(15 - h)$ cm.
Using Pythagoras' theorem on the right-angled triangle formed by the radius, the water surface, and the vertical axis:
$x^2 + (15 - h)^2 = 15^2$
$x^2 + 225 - 30h + h^2 = 225$
$x^2 = 30h - h^2$

2. Express the water volume $W$ in terms of $h$ and compute the rate of change of depth $\frac{dh}{dt}$ at $h=3$ (as determined in the preceding part c(i)).

Substitute $x^2 = 30h - h^2$ into the volume formula:
$W = \frac{\pi h(3(30h - h^2) + h^2)}{6} = \frac{\pi h(90h - 2h^2)}{6} = \pi\left(15h^2 - \frac{1}{3}h^3\right)$
Differentiating with respect to $t$:
$\frac{dW}{dt} = \pi(30h - h^2)\frac{dh}{dt}$
Given $\frac{dW}{dt} = -20$ (since water is leaking), substitute $h = 3$:
$-20 = \pi(30(3) - 3^2)\frac{dh}{dt}$
$-20 = 81\pi \frac{dh}{dt} \implies \frac{dh}{dt} = -\frac{20}{81\pi}$

3. Differentiate the geometric relationship with respect to time $t$ to link $\frac{dx}{dt}$ and $\frac{dh}{dt}$.

Differentiating $x^2 = 30h - h^2$ implicitly with respect to $t$:
$2x \frac{dx}{dt} = (30 - 2h)\frac{dh}{dt}$

4. Substitute $h=3$ and the known rates to evaluate the rate of change of the radius $x$.

When $h = 3$:
$x^2 = 30(3) - 3^2 = 81 \implies x = 9$ (since $x > 0$)

Substitute $x = 9$ and $\frac{dh}{dt} = -\frac{20}{81\pi}$ into the differentiated equation:
$2(9) \frac{dx}{dt} = (30 - 2(3))\left(-\frac{20}{81\pi}\right)$
$18 \frac{dx}{dt} = 24\left(-\frac{20}{81\pi}\right) = -\frac{480}{81\pi}$
$\frac{dx}{dt} = -\frac{480}{18 \times 81\pi} = -\frac{80}{243\pi}$