[ 底面半径为 $r$ 且高为 $h$ 的圆锥，其体积公式为 $V = \frac{1}{3}\pi r^2 h$。 ] 一种用于饮水机的一次性纸杯呈圆锥形，由一张半径为 8 cm、厚度可忽略不计的圆形纸片制成。从纸片上剪下一个扇形，如图1所示。然后将剩余的纸片折叠成一个母线长为 8 cm、底面半径为 $r$ cm、高为 $h$ cm 的圆锥，如图2所示。

1. Express the square of the base radius $r^2$ in terms of the height $h$.

\text{By Pythagoras' theorem, } r^2 + h^2 = 8^2 \implies r^2 = 64 - h^2

2. Substitute $r^2$ into the volume formula to express $V$ as a function of $h$.

V = \frac{1}{3}\pi r^2 h \
V = \frac{1}{3}\pi (64 - h^2) h \
V = \frac{1}{3}\pi (64h - h^3)

3. Differentiate the volume $V$ with respect to $h$.

\frac{\text{d}V}{\text{d}h} = \frac{1}{3}\pi (64 - 3h^2)

4. Set the derivative to zero to find the height that gives the stationary point.

\text{For stationary points, } \frac{\text{d}V}{\text{d}h} = 0: \
\frac{1}{3}\pi (64 - 3h^2) = 0 \
3h^2 = 64 \
h^2 = \frac{64}{3} \
h = \frac{8}{\sqrt{3}} \quad (\text{since } h > 0)

5. Verify the stationary point is a maximum using the second derivative test.

\frac{\text{d}^2V}{\text{d}h^2} = \frac{1}{3}\pi (-6h) = -2\pi h \
\text{When } h = \frac{8}{\sqrt{3}}, \frac{\text{d}^2V}{\text{d}h^2} = -2\pi \left(\frac{8}{\sqrt{3}}\right) < 0 \text{, so } V \text{ is a maximum.}

6. Substitute the optimal height back into the volume formula to find the exact maximum volume.

\text{Max } V = \frac{1}{3}\pi \left(64\left(\frac{8}{\sqrt{3}}\right) - \left(\frac{8}{\sqrt{3}}\right)^3\right) \
= \frac{1}{3}\pi \left(\frac{512}{\sqrt{3}} - \frac{512}{3\sqrt{3}}\right) \
= \frac{1}{3}\pi \left(\frac{1024}{3\sqrt{3}}\right) \
= \frac{1024\pi}{9\sqrt{3}} \
= \frac{1024\sqrt{3}\pi}{27}

1. Re-establish the geometric relationship and the condition for maximum volume from part (i).

By Pythagoras' theorem on the cone with slant height 8 cm:
$r^2 + h^2 = 8^2 \implies r^2 = 64 - h^2$
Expressing the volume $V$ in terms of $h$:
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (64 - h^2)h = \frac{1}{3}\pi (64h - h^3)$
For maximum volume, $\frac{\text{d}V}{\text{d}h} = 0$:
$\frac{\text{d}}{\text{d}h} \left[ \frac{1}{3}\pi (64h - h^3) \right] = 0$
$\frac{1}{3}\pi (64 - 3h^2) = 0 \implies 3h^2 = 64$

2. Substitute the condition back into the geometric relationship to show the result.

Substitute $64 = 3h^2$ into the expression for $r^2$:
$r^2 = 64 - h^2$
$r^2 = 3h^2 - h^2$
$r^2 = 2h^2$
Since $r$ and $h$ are positive lengths, taking the square root gives:
$r = \sqrt{2}h \text{ (shown)}$

1. Establish the relationship between the radius of the water surface $x$ and the water height $y$.

Let $y$ be the vertical height of the water in the cone.
From part (ii), we know that the dimensions of the given cone satisfy $r = \sqrt{2}h$.
By similar triangles, the water forms a smaller cone with dimensions proportional to the full cone, so the ratio of radius to height is constant:
$\frac{x}{y} = \frac{r}{h} = \sqrt{2}$
$x = \sqrt{2}y$

2. Formulate an equation for the volume of the water $V$ purely in terms of its height $y$.

The volume of the water $V$ at any given time is:
$V = \frac{1}{3}\pi x^2 y$
Substitute $x = \sqrt{2}y$:
$V = \frac{1}{3}\pi (\sqrt{2}y)^2 y$
$V = \frac{1}{3}\pi (2y^2) y$
$V = \frac{2}{3}\pi y^3$

3. Calculate the volume and height of the water at $t = 6$ seconds.

Since water flows in at a constant rate of $3\pi\text{ cm}^3\text{ per second}$, the volume of water after $6$ seconds is:
$V = 3\pi \times 6 = 18\pi$
Equating this to the volume formula to find the instantaneous height $y$:
$\frac{2}{3}\pi y^3 = 18\pi$
$y^3 = 18 \times \frac{3}{2}$
$y^3 = 27$
$y = 3$

4. Use the chain rule to find the rate of change of the water height.

Differentiating $V$ with respect to time $t$ using the chain rule:
$\frac{dV}{dt} = \frac{dV}{dy} \times \frac{dy}{dt}$
$3\pi = \left( \frac{d}{dy}\left[\frac{2}{3}\pi y^3\right] \right) \frac{dy}{dt}$
$3\pi = 2\pi y^2 \frac{dy}{dt}$
Substitute the instantaneous height $y = 3$:
$3\pi = 2\pi (3)^2 \frac{dy}{dt}$
$3\pi = 18\pi \frac{dy}{dt}$
$\frac{dy}{dt} = \frac{3\pi}{18\pi} = \frac{1}{6}$