A popular toy company is designing a new water play feature for children. The toy consists of a cylindrical water container that will hold water for various playful activities. This cylindrical container is designed to be inscribed within a fixed, rigid hemispherical shell made of durable plastic of negligible thickness. [Figure: cross-sectional diagram below shows a semicircle of radius $2a$ with a rectangle (shaded) inscribed inside it. The rectangle is symmetric about the vertical diameter. The half-width from the centre to the left vertical side of the rectangle is labelled $r\ \text{cm}$. The full base width of the hemisphere is labelled $2a\ \text{cm}$.] The shaded region in the diagram above shows the cross-sectional view of the upright cylindrical container that is inscribed in a hemisphere with fixed radius $2a\ \text{cm}$.

1. Express the height of the cylinder using Pythagoras' theorem and substitute it into the volume formula.

Let $h$ be the height of the cylindrical container.
From the cross-sectional diagram, the top corner of the rectangle touches the semicircle, forming a right-angled triangle with the radius of the hemisphere $2a$ as the hypotenuse.
By Pythagoras' theorem:
$h^2 + r^2 = (2a)^2$
$h = \sqrt{4a^2 - r^2}$
The volume of the cylinder is given by $V = \pi r^2 h$.
Substituting the expression for $h$, we get:
$V = \pi r^2 \sqrt{4a^2 - r^2}\ \text{(shown)}$

1. Find the radius $r$ of the container at the instant when the height $h = 80\ \text{cm}$.

Since the height of the container is given by $h = \sqrt{4a^2 - r^2}$, substitute $a = 50$ and $h = 80$:
$80 = \sqrt{4(50)^2 - r^2}$
$80 = \sqrt{10000 - r^2}$
$80^2 = 10000 - r^2$
$6400 = 10000 - r^2$
$r^2 = 3600 \implies r = 60 \text{ (since } r > 0)$

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

From part (a), $V = \pi r^2 \sqrt{4a^2 - r^2}$. Substituting $a = 50$, we have $V = \pi r^2 \sqrt{10000 - r^2}$.
Differentiating $V$ with respect to $r$ using the product and chain rules:
$\frac{dV}{dr} = \pi \left[ \left( \frac{d}{dr}(r^2) \right) \sqrt{10000 - r^2} + r^2 \left( \frac{d}{dr}(10000 - r^2)^{\frac{1}{2}} \right) \right]$
$\frac{dV}{dr} = \pi \left[ 2r(10000 - r^2)^{\frac{1}{2}} + r^2 \left( \frac{1}{2}(10000 - r^2)^{-\frac{1}{2}}(-2r) \right) \right]$
$\frac{dV}{dr} = \pi \left[ 2r\sqrt{10000 - r^2} - \frac{r^3}{\sqrt{10000 - r^2}} \right]$

3. Evaluate $\frac{dV}{dr}$ at $r = 60$.

Substitute $r = 60$ into the expression for $\frac{dV}{dr}$:
$\frac{dV}{dr} = \pi \left[ 2(60)\sqrt{10000 - 3600} - \frac{60^3}{\sqrt{10000 - 3600}} \right]$
$\frac{dV}{dr} = \pi \left[ 120(80) - \frac{216000}{80} \right]$
$\frac{dV}{dr} = \pi (9600 - 2700) = 6900\pi$

4. Use the connected rates of change formula to find $\frac{dr}{dt}$.

By the chain rule relating the rates of change over time:
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$
Given that water is pumped into the container at a rate of $100\pi\ \text{cm}^3\text{s}^{-1}$, we have $\frac{dV}{dt} = 100\pi$. Substituting our known values:
$100\pi = 6900\pi \times \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{100\pi}{6900\pi} = \frac{1}{69}$

1. Differentiate $V$ with respect to $r$ using the product and chain rules.

$V = \pi r^2(4a^2 - r^2)^{\frac{1}{2}}$
$\frac{dV}{dr} = \pi \left[ 2r(4a^2 - r^2)^{\frac{1}{2}} + r^2 \left( \frac{1}{2}(4a^2 - r^2)^{-\frac{1}{2}}(-2r) \right) \right]$
$\frac{dV}{dr} = \pi \left[ 2r\sqrt{4a^2 - r^2} - \frac{r^3}{\sqrt{4a^2 - r^2}} \right]$
$\frac{dV}{dr} = \frac{\pi \left[ 2r(4a^2 - r^2) - r^3 \right]}{\sqrt{4a^2 - r^2}}$
$\frac{dV}{dr} = \frac{\pi r(8a^2 - 3r^2)}{\sqrt{4a^2 - r^2}}$

2. Equate the derivative to zero to find the value of $r$ for the stationary volume.

For a maximum volume, $\frac{dV}{dr} = 0$.
Since $r > 0$, the denominator $\sqrt{4a^2 - r^2} \neq 0$ and $r \neq 0$.
$8a^2 - 3r^2 = 0$
$3r^2 = 8a^2$
$r^2 = \frac{8}{3}a^2$
$r = \sqrt{\frac{8}{3}}a = \frac{2\sqrt{6}}{3}a$

3. Justify that this value of $r$ gives a maximum volume using the first derivative test.

Using the first derivative test to examine the sign of $\frac{dV}{dr}$ around $r = \sqrt{\frac{8}{3}}a$:

The sign of $\frac{dV}{dr}$ is determined by the factor $(8a^2 - 3r^2)$ because $\frac{\pi r}{\sqrt{4a^2 - r^2}} > 0$.

• For $r < \sqrt{\frac{8}{3}}a$, we have $r^2 < \frac{8}{3}a^2 \implies 8a^2 - 3r^2 > 0$, so $\frac{dV}{dr} > 0$.
• At $r = \sqrt{\frac{8}{3}}a$, we have $\frac{dV}{dr} = 0$.
• For $r > \sqrt{\frac{8}{3}}a$, we have $r^2 > \frac{8}{3}a^2 \implies 8a^2 - 3r^2 < 0$, so $\frac{dV}{dr} < 0$.

Since the gradient changes from positive to negative, $V$ is a maximum when $r = \frac{2\sqrt{6}}{3}a$.

4. Calculate the exact maximum volume by substituting the optimal radius into the volume equation.

Substitute $r^2 = \frac{8}{3}a^2$ into $V = \pi r^2 \sqrt{4a^2 - r^2}$:
$V_{\text{max}} = \pi \left( \frac{8}{3}a^2 \right) \sqrt{4a^2 - \frac{8}{3}a^2}$
$V_{\text{max}} = \frac{8\pi a^2}{3} \sqrt{\frac{4}{3}a^2}$
$V_{\text{max}} = \frac{8\pi a^2}{3} \left( \frac{2a}{\sqrt{3}} \right)$
$V_{\text{max}} = \frac{16\pi a^3}{3\sqrt{3}} = \frac{16\sqrt{3}\pi}{9}a^3$

1. Determine the physical domain and the $r$-axis intercepts of the volume graph.

Since $r$ is the radius of the inscribed cylinder within a hemisphere of base radius $2a$, the valid physical range for $r$ is $0 \le r \le 2a$.
Setting $V = 0$ to find the intercepts:
$\pi r^2\sqrt{4a^2 - r^2} = 0$
$r = 0 \quad \text{or} \quad r = 2a$
The curve must start at the origin $(0, 0)$ and meet the $r$-axis again at $(2a, 0)$.

2. Incorporate the maximum point from part (c) and describe the shape of the sketch.

From the results found in part (c), the curve has a maximum volume at $r = a\sqrt{\frac{8}{3}} = \frac{2\sqrt{6}}{3}a$, with $V_{max} = \frac{16\sqrt{3}\pi a^3}{9}$.
Because $V \ge 0$ for all valid $r$ and there is exactly one turning point (a maximum), the graph is a smooth, hill-shaped (inverted U) curve connecting $(0,0)$ to $(2a,0)$ smoothly through the maximum point.