[tex]\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad\qquad V=16000\pi \implies 16000\pi=\pi r^2 h
\\\\\\
\cfrac{16000\pi }{\pi r^2}=h\implies \boxed{\cfrac{16000}{r^2}=h}\\\\
-------------------------------\\\\
\textit{surface area of a cylinder}\\\\
s=2\pi r^2+2\pi r h\qquad \qquad s(r)=2\pi r^2+2\pi r\left( \frac{16000}{r^2} \right)
\\\\\\
\boxed{s(r)=2\pi r^2+32000r^{-1}}
\\\\\\
\cfrac{ds}{dr}=4\pi r-\cfrac{32000\pi }{r^2}\implies \cfrac{ds}{dr}=\cfrac{4\pi r^3-32000\pi }{r^2}[/tex]
now, we could get a critical point from zeroing the denominator, however, in this case, we only get 0, and a radius of 0, gives us no volume and thus no cylinder, so, is not feasible.
now, let's get the other critical values by zeroing out derivative.
[tex]\bf 0=4\pi r^3-32000\pi\implies 32000\pi =4\pi r^3\implies \sqrt[3]{\cfrac{32000\pi }{4\pi }}=r
\\\\\\
\sqrt[3]{8000}=r\implies \boxed{20=r}[/tex]
now, doing a first-derivative test on the left and right sides of 20, say 19.999 and 20.001, check the picture below.
