A right circular cone has base radius \(r\), height \(h\) and
slant length \(\ell\). Its volume \(V\), and the area \(A\) of its curved
surface, are given by
\[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \]
Given that \(A\) is fixed and \(r\) is chosen so that \(V\) is at its stationary value, show that \(A^2 = 3\pi^2r^4\) and that
\(\ell =\sqrt3\,r\).
Given, instead, that \(V\) is fixed and \(r\) is chosen so that \(A\) is at its stationary value, find \(h\) in terms of \(r\).
Solution:
Given \(A\) is fixed, and \(h^2 + r^2 = \ell^2\), we can look at
\begin{align*}
&& V^2 &= \frac19 \pi^2 r^4 h^2 \\
&&&= \frac19\pi^2r^4(\ell^2 - r^2) \\
&&&= \frac19\pi^2 r^4\left (\frac{A^2}{\pi^2r^2} - r^2 \right) \\
&&&= \frac{A^2r^2 - \pi^2r^6}{9}
\end{align*}
Differentiating wrt to \(r\) we find that \(2rA^2-6\pi^2 r^5 = 0\) or hence \(A^2 = 3\pi^2 r^4 \Rightarrow A = \sqrt{3}\pi r^2\).
Therefore \(\sqrt{3}\pi r^2 = \pi r \ell \Rightarrow \ell = \sqrt{3}r\).
Supposing \(V\) is fixed, then
\begin{align*}
&& A^2 &= \pi^2 r^2\ell^2 \\
&&&= \pi^2 r^2 (h^2+r^2) \\
&&&= \pi^2 r^2 \left ( \frac{9V^2}{\pi^2r^4} + r^2 \right) \\
&&&= 9V^2r^{-2} + \pi^2r^4 \\
\end{align*}
Differentiating wrt to \(r\) we find \(-18V^2r^{-3} + 4\pi^2 r^3 = 0\) so \(V^2 = \frac{2\pi^2}{9}r^6\) or \(V = \frac{\sqrt{2}\pi}{3}r^3\), from which it follows: \(\frac{\sqrt{2}\pi}{3}r^3 = \frac13\pi r^2 h \Rightarrow h = \sqrt{2}r\)