[A sphere of radius $r$ has surface area $4\pi r^2$ and volume $\frac{4}{3}\pi r^3$.] A water fountain is to be constructed in the middle of Bishan East Park. It consists of a hemisphere with radius $r$ m joined to an open cylinder with radius $r$ m and height $h$ m (see diagram). The thickness of the fountain is negligible. It is given that the fountain, when filled to the brim, can hold a fixed volume $k$ m$^3$ of water.

1. Express the total volume of the fountain and make $h$ the subject of the formula.

The volume of the fountain is the sum of the volume of the hemisphere and the cylinder.
$V = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) + \pi r^2 h = k$
$k = \frac{2}{3}\pi r^3 + \pi r^2 h$
$\pi r^2 h = k - \frac{2}{3}\pi r^3$
$h = \frac{k - \frac{2}{3}\pi r^3}{\pi r^2}$

2. Formulate an expression for the total cost of painting the interior surface.

The interior surface to be painted consists of the curved surface of the hemisphere and the curved surface of the cylindrical wall.
Area of hemispherical surface $= \frac{1}{2}(4\pi r^2) = 2\pi r^2$
Area of cylindrical wall $= 2\pi r h$
Cost of painting hemispherical surface $= 3 \times 2\pi r^2 = 6\pi r^2$
Cost of painting cylindrical wall $= 2.50 \times 2\pi r h = 5\pi r h$
Total cost, $C = 6\pi r^2 + 5\pi r h$

3. Substitute the expression for $h$ into the cost equation and simplify to obtain the given result.

$C = 6\pi r^2 + 5\pi r \left( \frac{k - \frac{2}{3}\pi r^3}{\pi r^2} \right)$
$C = 6\pi r^2 + \frac{5(k - \frac{2}{3}\pi r^3)}{r}$
$C = 6\pi r^2 + \frac{5k}{r} - \frac{10}{3}\pi r^2$
$C = \left( 6 - \frac{10}{3} \right)\pi r^2 + \frac{5k}{r}$
$C = \left( \frac{18}{3} - \frac{10}{3} \right)\pi r^2 + \frac{5k}{r}$
$C = \frac{8}{3}\pi r^2 + \frac{5k}{r} \quad \text{(shown)}$

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

Given $C = \frac{8}{3}\pi r^2 + 5k r^{-1}$

$\frac{\mathrm{d}C}{\mathrm{d}r} = 2 \left(\frac{8}{3}\pi r\right) + 5k(-1)r^{-2}$
$\frac{\mathrm{d}C}{\mathrm{d}r} = \frac{16}{3}\pi r - \frac{5k}{r^2}$

2. Set the first derivative to zero and rearrange to solve for $r^3$.

For a stationary value, $\frac{\mathrm{d}C}{\mathrm{d}r} = 0$:

$\frac{16}{3}\pi r - \frac{5k}{r^2} = 0$
$\frac{16}{3}\pi r = \frac{5k}{r^2}$
Multiply both sides by $3r^2$:
$16\pi r^3 = 15k$
$r^3 = \frac{15k}{16\pi}$

3. Express $r$ in terms of $k$ and verify the nature of the stationary point using the second derivative test.

$r = \sqrt[3]{\frac{15k}{16\pi}}$

To verify it is a minimum, find the second derivative:
$\frac{\mathrm{d}^2C}{\mathrm{d}r^2} = \frac{16}{3}\pi - 5k(-2)r^{-3}$
$\frac{\mathrm{d}^2C}{\mathrm{d}r^2} = \frac{16}{3}\pi + \frac{10k}{r^3}$

Since volume $k > 0$ and radius $r > 0$, $\frac{\mathrm{d}^2C}{\mathrm{d}r^2} > 0$.
Therefore, $C$ is a minimum when $r = \sqrt[3]{\frac{15k}{16\pi}}$.

1. Determine the value of $r$ that minimizes $C$ when $k=50$.

For a minimum cost $C$, we use the condition $\frac{\mathrm{d}C}{\mathrm{d}r} = 0$ from the previous part:
$\frac{\mathrm{d}}{\mathrm{d}r} \left( \frac{8}{3}\pi r^2 + \frac{5k}{r} \right) = \frac{16}{3}\pi r - \frac{5k}{r^2} = 0$
Given $k = 50$:
$\frac{16}{3}\pi r = \frac{250}{r^2}$
$r^3 = \frac{750}{16\pi} = \frac{375}{8\pi}$
$r = \sqrt[3]{\frac{375}{8\pi}} \approx 2.4618$

2. Set up the volume equation for the fountain.

The total volume consists of the open cylinder and the hemisphere:
$V = \pi r^2 h + \frac{2}{3}\pi r^3$
Since the fountain holds a fixed volume $k = 50$, we have:
$\pi r^2 h + \frac{2}{3}\pi r^3 = 50$

3. Substitute the value of $r$ to solve for $h$.

Substituting $r^3 = \frac{375}{8\pi}$ and $r \approx 2.4618$:
$\pi (2.4618)^2 h + \frac{2}{3}\pi \left(\frac{375}{8\pi}\right) = 50$
$\pi (2.4618)^2 h + 31.25 = 50$
$\pi (2.4618)^2 h = 18.75$
$h = \frac{18.75}{\pi (2.4618)^2} \approx 0.9847$
Rounding to 3 significant figures:
$r \approx 2.46 \text{ m}, \quad h \approx 0.985 \text{ m}$

1. Determine the radius $r$ of the fountain when $k=50$ (carrying forward the condition for minimum cost from earlier parts).

From part (b), for the cost $C$ to be a minimum:
$\frac{\mathrm{d}C}{\mathrm{d}r} = \frac{16}{3}\pi r - \frac{5k}{r^2} = 0 \implies r^3 = \frac{15k}{16\pi}$
Substitute $k = 50$ (given in part (c)):
$r = \sqrt[3]{\frac{15(50)}{16\pi}} = \sqrt[3]{\frac{375}{8\pi}} \approx 2.4618 \text{ m}$

2. Set up the connected rates of change equation relating the volume of the cylindrical part to the water level.

Since the fountain is filled to the brim and the leak is at the joint below the cylinder, the water level decreases within the cylindrical section first.
Let $y$ be the water level in the cylinder. The volume of water in this part is:
$V = \pi r^2 y$
Differentiating with respect to time $t$ gives:
$\frac{\mathrm{d}V}{\mathrm{d}t} = \pi r^2 \frac{\mathrm{d}y}{\mathrm{d}t}$

3. Substitute the given rate of leak and the calculated area to find the rate at which the water level is decreasing.

We are given that the water leaks out at $0.002 \text{ m}^3\text{/min}$, so $\frac{\mathrm{d}V}{\mathrm{d}t} = -0.002$.
$-0.002 = \pi (2.4618)^2 \frac{\mathrm{d}y}{\mathrm{d}t}$
$-0.002 = 19.040 \frac{\mathrm{d}y}{\mathrm{d}t}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{-0.002}{19.040} \approx -0.00010504 \text{ m/min}$
The rate at which the water level is decreasing is $0.000105 \text{ m/min}$ (or $1.05 \times 10^{-4} \text{ m/min}$ to 3 s.f.).