A cone-shaped cup is made of paper of negligible thickness to hold $20\pi\text{ cm}^3$ of liquid. The open inverted cone has radius $r\ \text{cm}$ and height $h\ \text{cm}$ as shown in the diagram above. The external surface area of the cup is denoted by $A\ \text{cm}^2$. The manufacturer wants to reduce the cost of production by minimizing the value of $A$.

1. Equate the formula for the volume of a cone to the given volume.

$\frac{1}{3}\pi r^2 h = 20\pi$

2. Make $h$ the subject of the equation.

$\frac{1}{3} r^2 h = 20 \implies r^2 h = 60 \implies h = \frac{60}{r^2}$

1. Express $A^2$ entirely in terms of $r$.

From part (a), the volume is $V = \frac{1}{3}\pi r^2 h = 20\pi$, which gives $h = \frac{60}{r^2}$.
The external surface area is $A = \pi r l = \pi r \sqrt{r^2 + h^2}$.
Squaring both sides:
$A^2 = \pi^2 r^2 (r^2 + h^2)$
Substituting $h = \frac{60}{r^2}$:
$A^2 = \pi^2 r^2 \left(r^2 + \left(\frac{60}{r^2}\right)^2\right)$
$A^2 = \pi^2 r^2 \left(r^2 + \frac{3600}{r^4}\right)$
$A^2 = \pi^2 \left(r^4 + 3600r^{-2}\right)$

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

Differentiating both sides with respect to $r$:
$\frac{d}{dr}(A^2) = \frac{d}{dr}\left[\pi^2(r^4 + 3600r^{-2})\right]$
$2A \frac{dA}{dr} = \pi^2 \left(4r^3 + 3600(-2)r^{-3}\right)$
$2A \frac{dA}{dr} = \pi^2 \left(4r^3 - \frac{7200}{r^3}\right)$

3. Divide by 2 to obtain the final expression.

Dividing the entire equation by 2:
$A \frac{dA}{dr} = \pi^2 \left(2r^3 - \frac{3600}{r^3}\right)$

1. Set the given derivative to zero to find the stationary value of $r$.

For a stationary value, $\frac{dA}{dr} = 0$.
Since $A > 0$ (external surface area), we have $A \frac{dA}{dr} = 0$.
Using the result from part (b):
$\pi^2\left(2r^3 - \frac{3600}{r^3}\right) = 0$
$2r^3 = \frac{3600}{r^3}$
$r^6 = 1800$
$r = \sqrt[6]{1800}$

2. Use the first derivative test to prove that this value of $r$ gives a minimum area.

We test the sign of $\frac{dA}{dr}$ around the exact value $r = \sqrt[6]{1800} \approx 3.49$.
Since $A$ is positive, the sign of $\frac{dA}{dr}$ is determined by the term $\left(2r^3 - \frac{3600}{r^3}\right)$.

• For $r = 3 < \sqrt[6]{1800}$: $2(3)^3 - \frac{3600}{3^3} = 54 - 133.33 < 0$, so $\frac{dA}{dr} < 0$.
• For $r = 4 > \sqrt[6]{1800}$: $2(4)^3 - \frac{3600}{4^3} = 128 - 56.25 > 0$, so $\frac{dA}{dr} > 0$.

Since the gradient of $A$ changes from negative to positive at $r = \sqrt[6]{1800}$, $A$ is a minimum.

3. Express the ratio $\frac{r}{h}$ in terms of $r$ using the volume formula.

From the volume of the cone given in the question stem, $V = 20\pi$:
$\frac{1}{3}\pi r^2 h = 20\pi$
$r^2 h = 60 \implies h = \frac{60}{r^2}$

The ratio is therefore:
$\frac{r}{h} = \frac{r}{\frac{60}{r^2}} = \frac{r^3}{60}$

4. Evaluate the exact ratio using the stationary value found.

Since $r^6 = 1800$ at the minimum, we can find $r^3$ by taking the square root:
$r^3 = \sqrt{1800} = \sqrt{900 \times 2} = 30\sqrt{2}$

Substitute this into our expression for the ratio:
$\frac{r}{h} = \frac{30\sqrt{2}}{60} = \frac{1}{2}\sqrt{2}$

(This matches the required form $k\sqrt{2}$ with $k = \frac{1}{2}$.)

1. Find the minimum ratio $\frac{r}{h}$ using the result from part (c).

From part (a), $h = \frac{60}{r^2}$.
For minimum $A$, $A\frac{dA}{dr} = 0 \implies 2r^3 - \frac{3600}{r^3} = 0 \implies r^6 = 1800 \implies r^3 = 30\sqrt{2}$.
Thus, the required ratio is $\frac{r}{h} = \frac{r}{60/r^2} = \frac{r^3}{60} = \frac{30\sqrt{2}}{60} = \frac{\sqrt{2}}{2}$.

2. Formulate the volume of water $V$ in terms of its depth $x$.

Let the depth of water be $x$ and the radius of the water's surface be $R$.
By similar triangles, the ratio of radius to depth is constant:
$\frac{R}{x} = \frac{r}{h} = \frac{\sqrt{2}}{2} \implies R = \frac{\sqrt{2}}{2}x$
The volume of water $V$ is:
$V = \frac{1}{3}\pi R^2 x = \frac{1}{3}\pi \left(\frac{\sqrt{2}}{2}x\right)^2 x = \frac{\pi}{6}x^3$

3. Apply the chain rule to relate the rates of change of volume and depth.

Differentiating $V$ with respect to time $t$:
$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = \frac{\pi}{2}x^2 \frac{dx}{dt}$

4. Substitute the given values to find the rate of decrease of the depth.

The water leaks out at $3\text{ cm}^3\text{/s}$, so $\frac{dV}{dt} = -3$.
At the instant when $x = 2$:
$-3 = \frac{\pi}{2}(2)^2 \frac{dx}{dt}$
$-3 = 2\pi \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{3}{2\pi}$
Since $\frac{dx}{dt}$ is negative, the rate of decrease of the depth is $\frac{3}{2\pi}\text{ cm/s}$.