[A sphere of radius $r$ has surface area $4\pi r^2$ and volume $\dfrac{4}{3}\pi r^3$.] A water fountain is to be constructed in the middle of Bishan East Park. It consists of a hemisphere with radius $r$ joined to an open cylinder with radius $r$ and height $h$ (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 $k$ 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 $1.05 \times 10^{-4} \text{ m/min}$).