# A cylindrical can, open at the top, is to hold 470cm^{3} of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent pi. Radius = ? Height = ?

A cylindrical can, open at the top, is to hold $470c{m}^{3}$ of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent pi.
Radius $=?$
Height $=?$
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coffentw
Given:
$V=470c{m}^{3}$
We have to find height and radius
$V=\pi {r}^{2}h=470$
$h-\frac{470}{\pi {r}^{2}}$
Plug this into the surface area equation
$SA=\pi {r}^{2}+2\pi rh$
$=\pi {r}^{2}+2\pi r\left(\frac{470}{\pi {r}^{2}}\right)$
$=\pi {r}^{2}+\frac{940}{r}$
Differentiate SA and set to 0 solve for r
$\frac{dSA}{dr}=2\pi r-\frac{940}{{r}^{2}}=0$
$2\pi r=\frac{940}{{r}^{2}}$
${r}^{3}=\frac{940}{2}x$
${r}^{3}=\frac{470}{\pi }$
$r=\left(\frac{470}{\pi }{\right)}^{\frac{1}{3}}$
$=5.30cm$
Now, we find h:
$h=\frac{470}{\pi {r}^{2}}$
$=\frac{470}{\pi \left(5.30{\right)}^{2}}$
$=5.32cm$