A closed right circular cone of base radius $r$ and height $h$ has a fixed total surface area $k\pi$ units$^2$, where $k$ is a positive constant. Show that $r^2 = \frac{k^2}{(h^2 + 2k)}$. Given that $r$ and $h$ vary, find the greatest possible volume of the cone, giving your answer in exact form in terms of $k$. [The volume of a cone of base radius $r$ and height $h$ is given by $V = \frac{1}{3}\pi r^2 h$ and the curved surface area of the cone is given by $A = \pi r l$, where $l$ is the slant height.]

1. Set up the equation for the total surface area and simplify.

$\pi r^2 + \pi r l = k\pi$
$r^2 + r\sqrt{r^2 + h^2} = k$

2. Rearrange the equation and square both sides to eliminate the square root.

$r\sqrt{r^2 + h^2} = k - r^2$
$(r\sqrt{r^2 + h^2})^2 = (k - r^2)^2$
$r^2(r^2 + h^2) = k^2 - 2kr^2 + r^4$
$r^4 + r^2 h^2 = k^2 - 2kr^2 + r^4$

3. Solve for $r^2$ to obtain the required identity.

$r^2 h^2 = k^2 - 2kr^2$
$r^2 h^2 + 2kr^2 = k^2$
$r^2(h^2 + 2k) = k^2$
$r^2 = \frac{k^2}{h^2 + 2k} \quad \text{(shown)}$

4. Express the volume of the cone in terms of a single variable $h$.

$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi \left(\frac{k^2}{h^2 + 2k}\right) h = \frac{\pi k^2 h}{3(h^2 + 2k)}$

5. Differentiate the volume $V$ with respect to $h$ using the quotient rule.

$\frac{dV}{dh} = \frac{\pi k^2}{3} \left[ \frac{(h^2 + 2k)(1) - h(2h)}{(h^2 + 2k)^2} \right]$
$\frac{dV}{dh} = \frac{\pi k^2}{3} \left[ \frac{h^2 + 2k - 2h^2}{(h^2 + 2k)^2} \right] = \frac{\pi k^2(2k - h^2)}{3(h^2 + 2k)^2}$

6. Set the derivative to zero to find the stationary point and verify it gives a maximum.

For a stationary point, $\frac{dV}{dh} = 0$:
$2k - h^2 = 0 \implies h^2 = 2k$
$h = \sqrt{2k} \quad \text{(since } h > 0\text{)}$
Using the first derivative test:

• For $0 < h < \sqrt{2k}$, $2k - h^2 > 0 \implies \frac{dV}{dh} > 0$
• For $h > \sqrt{2k}$, $2k - h^2 < 0 \implies \frac{dV}{dh} < 0$
  Thus, $V$ reaches a maximum when $h = \sqrt{2k}$.

7. Substitute $h = \sqrt{2k}$ into the volume expression to calculate the greatest possible volume.

$V_{\text{max}} = \frac{\pi k^2 (\sqrt{2k})}{3((\sqrt{2k})^2 + 2k)}$
$V_{\text{max}} = \frac{\pi k^2 \sqrt{2k}}{3(2k + 2k)} = \frac{\pi k^2 \sqrt{2k}}{12k} = \frac{\pi k \sqrt{2k}}{12}$