A rocket-shaped solid consists of a right circular cylinder of length \(l\) and radius \(r\) attached at one end to the base of a right circular cone of base radius \(r\) and height \(\frac{3}{4}r\). If the total volume of the solid is to be \(\quantity{14\pi}{cm^3}\), find its least possible surface area.
The volume of the rocket-shaped solid is the sum of the volumes of the cylinder and the cone.
The volume of the cylinder is \(\pi r^2 l\), and the volume of the cone is \(\frac{1}{3} \pi r^2 \left( \frac{3}{4} r \right)\), or \(\frac{1}{4}\pi r^3\).
From the question, we must have that \[\begin{equation*} 14\pi = \pi r^2 l + \frac{1}{4}\pi r^3 = \pi r^2 \left( l + \frac{r}{4} \right) \iff l = \frac{14}{r^2} - \frac{r}{4}. \end{equation*}\]The surface area of the rocket-shaped solid does not include the base of the cone or the top of the cylinder.
The contribution from the cylinder is thus its bottom disc and curved surface, while the contribution from the cone is just its sloped surface.
For the cylinder, the area of the bottom disc is \(\pi r^2\), while the area of the curved surface is \(2\pi r l\), since when uncurled, it’s a rectangle with side lengths \(2\pi r\) and \(l\).
The area of the sloped part of the cone is \(\pi r\) times the slope length; by Pythagoras’ theorem, this length is \[\begin{equation*} \sqrt{ r^2 + \left(\frac{3}{4}r \right)^2 } = \sqrt{ \frac{25}{16} r^2 } = \frac{5}{4} r. \end{equation*}\] Thus, the surface area equals \[\begin{align*} S(r)=\pi r^2 + 2 \pi r l + \pi r \frac{5}{4} r &= \pi r^2 \left( 1 + \frac{5}{4} \right) + 2 \pi r \left( \frac{14}{r^2} - \frac{r}{4} \right) \\ &= \frac{9}{4} \pi r^2 + \frac{28\pi}{r} - \frac{\pi}{2} r^2 \\ &= \frac{7}{4} \pi r^2 + \frac{28\pi}{r}. \end{align*}\] We are to minimise \(S(r)\) with respect to \(r\). Differentiating and setting the derivative equal to zero, \[\begin{equation*} \frac{dS}{dr} = \frac{7}{2} \pi r - \frac{28\pi}{r^2} = 0 \iff r^3 = 8 \iff r = 2. \end{equation*}\]This is the unique stationary point for \(S(r)\), and it must be a minimum, since \(S(r) \to \infty\) if \(r \to 0\) or \(r \to \infty\).
The least possible surface area is thus \[\begin{equation*} S(2) = \frac{7}{4} \pi \times 2^2 + \frac{28\pi}{2} = 7\pi + 14\pi = \quantity{21\pi}{cm^2}. \end{equation*}\]
