Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 38

Find the height and radius of the cup that will use the smallest amount of paper of a cone shaped paper drinking cup that is to be made to hold 27cm$^3$ of water.
Recall that the volume of the cone is $\displaystyle V = \frac{1}{3} \pi r^3 h = 27$, so...
$\displaystyle h = \frac{81}{\pi r^2}$

And, the surface area of the cone is equal to $s = \pi r \ell$



By Pythagorean Theorem,
$\ell = \sqrt{r^2 + h^2}$
So, $ S = \pi r \left(\sqrt{r^2 + h^2}\right)$

Substituting the value of $h$, we get...
$\displaystyle S = \pi r \sqrt{r^2 + \left( \frac{81}{\pi r^2} \right)^2}$

We can rewrite $S$ as...

$
\begin{equation}
\begin{aligned}
S &= \sqrt{\pi^2 r^2 \left( r^2 + \left( \frac{81}{\pi r^2} \right)^2 \right)}\\
\\
S &= \sqrt{\pi^2 + r^4 + \frac{81^2}{r^2}}
\end{aligned}
\end{equation}
$


Taking the derivative of $S$ with respect to $r$, we get...

$
\begin{equation}
\begin{aligned}
S' &= \frac{4\pi^2 r^3 + 81^2 \left( \frac{-2r}{r^4} \right)}{2 \sqrt{\pi^2 r^4 + \frac{81^2}{r^2}}}\\
\\
S' &= \frac{4\pi^2 r^3 + 81^2 \left( \frac{-2}{8r^3} \right)}{2 \sqrt{\pi^2 r^4 + \frac{81^2}{r^2}}}
\end{aligned}
\end{equation}
$


when $S' =0$

$
\begin{equation}
\begin{aligned}
0 &= 4\pi^2 r^3 + 81^2 \left( \frac{-2}{r^3} \right)\\
\\
\frac{13122}{r^3} &= 4 \pi^2 + r^3\\
\\
r &= \sqrt[6]{\frac{6561}{2\pi^2}} = 2.6319\text{cm}

\end{aligned}
\end{equation}
$


So when $r = 2.6319$cm, then
$\displaystyle h = \frac{81}{\pi r^2} = \frac{81}{\pi(2.6319)^2} = 3.7221$cm

Therefore, the radius and height of the cup that will use the smallest amount of paper will be $r = 2.6319$cm and $ h = 3.7221$cm respectively.