A cone of height 12 cm and base radius 5 cm is placed over a cylinder of radius $r$ cm and height $h$ cm. The cone is in contact with the cylinder along the cylinder's upper rim. The diagram shows a vertical cross-section of the cone and the cylinder.

1. Identify similar triangles from the vertical cross-section

Let the vertex of the cone be $V$, the center of the base be $O$, and a point on the base circumference be $A$. The triangle $VOA$ is a right-angled triangle with height $VO = 12$ and base $OA = 5$.
The cylinder's upper rim touches the cone. Let the point of contact be $B$, and let the projection of $B$ onto the central axis $VO$ be $C$. The triangle $VCB$ is similar to triangle $VOA$.

2. Set up the ratio of corresponding sides to find $h$ in terms of $r$

The height of the small triangle $VCB$ is $12 - h$, and its base is $r$.
Since $\triangle VCB \sim \triangle VOA$, the ratio of height to base is constant:
$\frac{12 - h}{r} = \frac{12}{5}$
$12 - h = \frac{12}{5}r$
$h = 12 - \frac{12}{5}r$

1. Find $h$ in terms of $r$ using similar triangles (from part i)

By similar triangles in the cross-section:
$\frac{12 - h}{12} = \frac{r}{5}$
$12 - h = \frac{12r}{5}$
$h = 12 - \frac{12}{5}r$

2. Substitute $h$ into the formula for the volume of a cylinder

The volume of the cylinder is $V = \pi r^2 h$.
Substituting $h$:
$V = \pi r^2 \left(12 - \frac{12}{5}r\right)$

3. Expand the expression to show the required result

$V = 12\pi r^2 - \frac{12}{5}\pi r^3$ (Shown)

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

$\frac{dV}{dr} = 24\pi r - \frac{36}{5}\pi r^2$

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

For maximum $V$, $\frac{dV}{dr} = 0$
$24\pi r - \frac{36}{5}\pi r^2 = 0$
$12\pi r \left(2 - \frac{3}{5}r\right) = 0$
Since $r > 0$, $2 - \frac{3}{5}r = 0 \implies r = \frac{10}{3}$

3. Verify that this value of $r$ gives a maximum volume

$\frac{d^2V}{dr^2} = 24\pi - \frac{72}{5}\pi r$
When $r = \frac{10}{3}$, $\frac{d^2V}{dr^2} = 24\pi - \frac{72}{5}\pi \left(\frac{10}{3}\right) = 24\pi - 48\pi = -24\pi$
Since $\frac{d^2V}{dr^2} < 0$, $V$ is a maximum when $r = \frac{10}{3}$.

4. Calculate the maximum value of $V$

Maximum $V = 12\pi \left(\frac{10}{3}\right)^2 - \frac{12}{5}\pi \left(\frac{10}{3}\right)^3$
$= 12\pi \left(\frac{100}{9}\right) - \frac{12}{5}\pi \left(\frac{1000}{27}\right)$
$= \frac{400\pi}{3} - \frac{800\pi}{9}$
$= \frac{1200\pi - 800\pi}{9}$
$= \frac{400\pi}{9} \approx 140$ cm$^3$ (to 3 s.f.)