[半径为 $r$ 的球体表面积为 $4\pi r^2$，体积为 $\dfrac{4}{3}\pi r^3$。] 碧山东公园正中央将建造一个喷水池。它由一个半径为 $r$ 的半球连接到一个半径为 $r$、高度为 $h$ 的无盖圆柱体组成（见图）。 喷水池壁的厚度忽略不计。已知当喷水池装满时，可以容纳固定体积为 $k$ $\text{m}^3$ 的水。

1. Express the total volume $k$ in terms of $r$ and $h$, and make $h$ the subject.

The total volume $k$ is the sum of the volume of the hemisphere and the volume of the cylinder:
$k = \frac{2}{3}\pi r^3 + \pi r^2 h$
$\pi r^2 h = k - \frac{2}{3}\pi r^3$
$h = \frac{k}{\pi r^2} - \frac{2r}{3}$

2. Formulate the total cost $C$ in terms of $r$ and $h$.

Area of the inner hemispherical surface $= 2\pi r^2$
Area of the inner cylindrical wall $= 2\pi rh$

Cost of painting hemispherical surface $= 3 \times 2\pi r^2 = 6\pi r^2$
Cost of painting cylindrical wall $= 2.50 \times 2\pi rh = 5\pi rh$

$C = 6\pi r^2 + 5\pi rh$

3. Substitute $h$ into the cost equation and simplify to obtain the final expression.

$C = 6\pi r^2 + 5\pi r \left( \frac{k}{\pi r^2} - \frac{2r}{3} \right)$
$C = 6\pi r^2 + \frac{5k}{r} - \frac{10}{3}\pi r^2$
$C = \left(6 - \frac{10}{3}\right)\pi r^2 + \frac{5k}{r}$
$C = \frac{8}{3}\pi r^2 + \frac{5k}{r} \text{ (Shown)}$

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

$\frac{\mathrm{d}C}{\mathrm{d}r} = \frac{16}{3}\pi r - \frac{5k}{r^2}$

2. Set the first derivative to zero to find the stationary point.

$\frac{16}{3}\pi r - \frac{5k}{r^2} = 0$

3. Solve for $r$ in terms of $k$.

$\frac{16}{3}\pi r = \frac{5k}{r^2}$\n$16\pi r^3 = 15k$\n$r^3 = \frac{15k}{16\pi}$\n$r = \sqrt[3]{\frac{15k}{16\pi}}$

4. Verify that this value of $r$ yields a minimum cost using the second derivative test.

$\frac{\mathrm{d}^2C}{\mathrm{d}r^2} = \frac{16}{3}\pi + \frac{10k}{r^3}$\nSince $k > 0$ and $r > 0$ for physical dimensions, $\frac{\mathrm{d}^2C}{\mathrm{d}r^2} > 0$.\nThus, the cost $C$ is a minimum at $r = \sqrt[3]{\frac{15k}{16\pi}}$.

1. Substitute $k = 50$ into the expression for $r$ obtained in part (b) to find its numerical value.

From part (b), we established that for a minimum cost $C$, $\frac{\mathrm{d}C}{\mathrm{d}r} = \frac{16}{3}\pi r - \frac{5k}{r^2} = 0$, which gives $r = \sqrt[3]{\frac{15k}{16\pi}}$.
Substituting $k = 50$:
$r = \sqrt[3]{\frac{15(50)}{16\pi}} = \sqrt[3]{\frac{375}{8\pi}}$
$r \approx 2.4618 \text{ m}$

2. Use the volume formula to find the corresponding value of $h$.

The total volume $k$ is the sum of the volumes of the cylinder and the hemisphere:
$k = \pi r^2 h + \frac{2}{3}\pi r^3$
Substituting $k = 50$ and the exact value of $r^3 = \frac{375}{8\pi}$:
$50 = \pi r^2 h + \frac{2}{3}\pi \left(\frac{375}{8\pi}\right)$
$50 = \pi r^2 h + 31.25$
$\pi r^2 h = 18.75$
$h = \frac{18.75}{\pi (2.4618)^2}$
$h \approx 0.98476 \text{ m}$
Rounding to 3 significant figures, we get:
$r = 2.46$, $h = 0.985$

1. Determine the cross-sectional area of the free surface using the value of $r$ from part (c).

From part (c), the radius is $r = \sqrt[3]{\frac{15(50)}{16\pi}} \approx 2.4618 \text{ m}$.
Since the fountain is filled to the brim, the water level is initially within the cylindrical section.
The volume of water $V$ when the water level is at height $y$ above the joint in the cylinder is given by $V = \frac{2}{3}\pi r^3 + \pi r^2 y$.
Differentiating with respect to $y$, the cross-sectional area is:
$\frac{\mathrm{d}V}{\mathrm{d}y} = \pi r^2 = \pi (2.4618)^2 \approx 19.040 \text{ m}^2$

2. Apply the chain rule to find the rate of decrease of the water level.

We are given the rate of water leaking, which represents the rate of change of volume with respect to time:
$\frac{\mathrm{d}V}{\mathrm{d}t} = -0.002 \text{ m}^3/\text{min}$
Using the chain rule, $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}y} \times \frac{\mathrm{d}y}{\mathrm{d}t}$:
$-0.002 = 19.040 \times \frac{\mathrm{d}y}{\mathrm{d}t}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{-0.002}{19.040} \approx -0.00010504 \text{ m/min}$
Thus, the rate at which the level of water is decreasing is $0.000105 \text{ m/min}$ (or $1.05 \times 10^{-4} \text{ m/min}$).