A closed container is made up of a cylinder of base radius $r$ cm and height $h$ cm, and a hemispherical top with the same radius $r$. It is inscribed within a fixed right circular cone of base radius 5 cm and height 12 cm, as shown in the diagram below.

1. Determine the slant height of the cone from its cross-section.

Let $V$ be the vertex of the cone, $O$ be the centre of its base, and $B$ be a point on the perimeter of the base.
Consider the vertical cross-section $\triangle VOB$, which is right-angled at $O$.
The vertical height $VO = 12$ and base radius $OB = 5$.
By Pythagoras' theorem, the slant edge is:
$VB = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}$

2. Set up a similarity relationship between the cone and the inscribed container.

Let $C$ be the centre of the hemispherical top. It lies on $VO$ such that $OC = h$, giving $VC = 12 - h$.
The hemisphere is tangent to the slant edge $VB$ at a point $T$. Thus, the radius to the point of tangency is perpendicular to the slant edge ($CT \perp VB$) and $CT = r$.
In $\triangle VTC$ and $\triangle VOB$:

• $\angle TVC = \angle OVB$ (common angle)
• $\angle VTC = \angle VOB = 90^\circ$
  Therefore, $\triangle VTC$ is similar to $\triangle VOB$.
  Using the ratio of corresponding sides:
  $\frac{CT}{OB} = \frac{VC}{VB}$
  $\frac{r}{5} = \frac{12 - h}{13}$

3. Rearrange to show the required expression for $h$ and determine the exact range of $r$.

$13r = 5(12 - h)$
$13r = 60 - 5h$
$5h = 60 - 13r$
$h = 12 - \frac{13}{5}r \quad \text{(shown)}$
For the container to be physically composed of a cylinder and a hemisphere, its dimensions must be strictly positive.
Since $h > 0$:
$12 - \frac{13}{5}r > 0$
$\frac{13}{5}r < 12 \implies r < \frac{60}{13}$
Since $r$ is a length, $r > 0$. Therefore, the exact range of possible values for $r$ is:
$0 < r < \frac{60}{13}$

1. Express the total volume $V$ of the closed container in terms of $r$.

The container consists of a cylinder and a hemisphere.
$V = V_{\text{cylinder}} + V_{\text{hemisphere}}$
$V = \pi r^2 h + \frac{2}{3}\pi r^3$
Substitute $h = 12 - \frac{13}{5}r$ from part (i):
$V = \pi r^2 \left(12 - \frac{13}{5}r\right) + \frac{2}{3}\pi r^3$
$V = 12\pi r^2 - \frac{13}{5}\pi r^3 + \frac{2}{3}\pi r^3$
$V = 12\pi r^2 + \left(-\frac{39}{15} + \frac{10}{15}\right)\pi r^3$
$V = 12\pi r^2 - \frac{29}{15}\pi r^3$

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

$\frac{dV}{dr} = 24\pi r - 3\left(\frac{29}{15}\pi\right)r^2$
$\frac{dV}{dr} = 24\pi r - \frac{29}{5}\pi r^2$

3. Set the derivative to zero and solve for the optimal value of $r$.

To find the stationary point, set $\frac{dV}{dr} = 0$:
$24\pi r - \frac{29}{5}\pi r^2 = 0$
$\pi r \left( 24 - \frac{29}{5}r \right) = 0$
Since $r > 0$ for a valid physical container, we reject $r = 0$:
$24 - \frac{29}{5}r = 0$
$\frac{29}{5}r = 24 \implies r = \frac{120}{29}$

4. Verify that this value of $r$ produces a maximum volume using the second derivative test.

$\frac{d^2V}{dr^2} = 24\pi - 2\left(\frac{29}{5}\pi\right)r$
$\frac{d^2V}{dr^2} = 24\pi - \frac{58}{5}\pi r$
Substitute $r = \frac{120}{29}$:
$\frac{d^2V}{dr^2} = 24\pi - \frac{58}{5}\pi \left(\frac{120}{29}\right)$
$\frac{d^2V}{dr^2} = 24\pi - 2(24)\pi = -24\pi$
Since $\frac{d^2V}{dr^2} < 0$, the volume is indeed maximized at $r = \frac{120}{29}$.