裱花袋呈圆锥形，半径为 20 cm，高为 30 cm。一位糕点师在裱花袋中装入糖霜，深度为 $x\text{ cm}$，半径为 $r\text{ cm}$。在裱花时，糖霜从裱花袋的尖端以 $80\text{ cm}^3/\text{s}$ 的恒定速率挤出。

1. Express the radius of the frosting, $r$, in terms of its depth, $x$, using similar triangles.

By similar triangles,
$\frac{r}{x} = \frac{20}{30}$
$r = \frac{2}{3}x$

2. Substitute $r$ into the formula for the volume of a cone and simplify.

The volume of the frosting is given by:
$V = \frac{1}{3}\pi r^2 x$
Substitute $r = \frac{2}{3}x$:
$V = \frac{1}{3}\pi \left(\frac{2}{3}x\right)^2 x$
$V = \frac{1}{3}\pi \left(\frac{4}{9}x^2\right) x$
$V = \frac{4}{27}\pi x^3 \text{ (Shown)}$

1. Find the depth $x$ when $r = 12$ using similar triangles.

By similar triangles,
$\frac{r}{x} = \frac{20}{30}$
$r = \frac{2}{3}x$
When $r = 12$,
$12 = \frac{2}{3}x \implies x = 18$

2. Differentiate the volume $V$ with respect to $x$.

$V = \frac{4}{27}\pi x^3$
$\frac{dV}{dx} = \frac{4}{27}\pi (3x^2) = \frac{4}{9}\pi x^2$

3. Apply the chain rule to relate the rates of change.

By the chain rule,
$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$

4. Substitute the known values to find the rate of change of the depth.

Since the frosting is squeezed out, the volume is decreasing, so $\frac{dV}{dt} = -80$.
$-80 = \frac{4}{9}\pi (18)^2 \times \frac{dx}{dt}$
$-80 = 144\pi \times \frac{dx}{dt}$
$\frac{dx}{dt} = -\frac{80}{144\pi} = -\frac{5}{9\pi}$

1. Calculate the value of $x$ when $r = 12$ using similar triangles.

From the dimensions of the cone, by similar triangles:
$\frac{r}{x} = \frac{20}{30} \implies r = \frac{2}{3}x \implies x = \frac{3}{2}r$
When $r = 12$, $x = \frac{3}{2}(12) = 18$.

2. Find the rate of change of depth, $\frac{dx}{dt}$, when $r = 12$ (result from part ii).

Given $V = \frac{4}{27}\pi x^3$, we have $\frac{dV}{dx} = \frac{4}{9}\pi x^2$.
When $x = 18$, $\frac{dV}{dx} = \frac{4}{9}\pi (18)^2 = 144\pi$.
Using the chain rule $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$ and given $\frac{dV}{dt} = -80$:
$-80 = 144\pi \times \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{80}{144\pi} = -\frac{5}{9\pi}$

3. Use the relationship between $r$ and $x$ to find $\frac{dr}{dt}$.

Since $r = \frac{2}{3}x$, differentiating with respect to $t$ gives:
$\frac{dr}{dt} = \frac{2}{3} \frac{dx}{dt}$
$\frac{dr}{dt} = \frac{2}{3} \left(-\frac{5}{9\pi}\right) = -\frac{10}{27\pi}$
$\approx -0.118\text{ cm/s}$ (to 3 s.f.)