[已知半径为 $r$、高为 $h$、母线长为 $l$ 的圆锥的侧面积为 $\pi r l$。]

1. Use the formula for the volume of a cone to express $h$ in terms of $r$.

$V = \frac{1}{3}\pi r^2 h = 50\pi$
$r^2 h = 150 \implies h = \frac{150}{r^2}$

2. Express the square of the curved surface area, $A^2$, in terms of $r$ and $h$.

$A = \pi r l$
Using Pythagoras' theorem for the slant height, $l^2 = r^2 + h^2$.
$A^2 = \pi^2 r^2 l^2 = \pi^2 r^2 (r^2 + h^2)$

3. Substitute the expression for $h$ to show the given formula for $A^2$.

$A^2 = \pi^2 r^2 \left( r^2 + \left(\frac{150}{r^2}\right)^2 \right)$
$A^2 = \pi^2 r^2 \left( r^2 + \frac{22500}{r^4} \right)$
$A^2 = \pi^2 r^4 + \frac{22500\pi^2}{r^2}$

4. Differentiate $A^2$ with respect to $r$.

Let $Z = A^2$. To find the least amount of material, we minimise $Z$ (since $A > 0$).
Rewrite $Z$ as $Z = \pi^2 r^4 + 22500\pi^2 r^{-2}$.
$\frac{dZ}{dr} = 4\pi^2 r^3 - 45000\pi^2 r^{-3}$
$\frac{dZ}{dr} = 4\pi^2 r^3 - \frac{45000\pi^2}{r^3}$

5. Equate the derivative to zero and solve for $r$.

For a stationary point, $\frac{dZ}{dr} = 0$:
$4\pi^2 r^3 - \frac{45000\pi^2}{r^3} = 0$
$4\pi^2 r^6 = 45000\pi^2$
$r^6 = 11250$
$r = \sqrt[6]{11250} \approx 4.7177$

6. Verify that this value of $r$ gives a minimum area using the second derivative test.

$\frac{d^2Z}{dr^2} = 12\pi^2 r^2 + 135000\pi^2 r^{-4} = 12\pi^2 r^2 + \frac{135000\pi^2}{r^4}$
Since $r > 0$, $\frac{d^2Z}{dr^2} > 0$. Thus, $Z$ (and therefore the curved surface area $A$) is a minimum.

7. Calculate the corresponding value of $h$.

$h = \frac{150}{r^2} = \frac{150}{(\sqrt[6]{11250})^2} = \frac{150}{\sqrt[3]{11250}} \approx 6.6943$
Therefore, to 3 significant figures, $r \approx 4.72$ cm and $h \approx 6.69$ cm.

1. Find the derivative of volume $V$ with respect to depth $h$.

$V = \frac{\pi h^3}{12}$
$\frac{dV}{dh} = \frac{\pi}{12}(3h^2) = \frac{\pi h^2}{4}$

2. Set up the connected rate of change equation and substitute the given values.

Using the chain rule, we have:
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$
The volume is decreasing at $3\text{ cm}^3\text{s}^{-1}$, so $\frac{dV}{dt} = -3$.
Substituting $h = 2$ and $\frac{dV}{dt} = -3$:
$-3 = \frac{\pi (2)^2}{4} \times \frac{dh}{dt}$

3. Solve for $\frac{dh}{dt}$ and state the rate of decrease.

$-3 = \pi \times \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{3}{\pi}$
Since $\frac{dh}{dt}$ is negative, the depth is decreasing at a rate of $\frac{3}{\pi}\text{ cm s}^{-1}$ (approx. $0.955\text{ cm s}^{-1}$).

1. Calculate the volume of water at the initial depth of $6\text{ cm}$ and final depth of $3\text{ cm}$.

$V_{\text{initial}} = \frac{\pi (6)^3}{12} = 18\pi$
$V_{\text{final}} = \frac{\pi (3)^3}{12} = \frac{27\pi}{12} = 2.25\pi$

2. Find the total change in volume and use the constant rate of volume decrease to find the time taken.

$\text{Change in volume} = 18\pi - 2.25\pi = 15.75\pi$
Since the volume decreases at a constant rate of $3\text{ cm}^3\text{s}^{-1}$:
$\text{Time taken} = \frac{\text{Change in volume}}{\text{Rate of decrease}} = \frac{15.75\pi}{3}$
$t = 5.25\pi \approx 16.5\text{ s (to 3 s.f.)}$