Originally Posted by

**redskies** A company prodces cylindrical cans.The volume of each can must be

$\displaystyle 16\pi(cm^3)$

The thickness of the metal will be the same no matter what the volume of the can.What should radius be to minimize amount of metal used?

Total surface area:

$\displaystyle 2\pi(r)(h)+2\pi(r)^2$

Volume :$\displaystyle \pi(r^2)(h)=16\pi$

Therefore :

$\displaystyle h=\frac{16}{r^2}$

Filling into Surface area:

$\displaystyle 2\pi(r)(\frac{16}{r^2})+2\pi(r^2)$

$\displaystyle 32\pi(r^{-1})+2\pi(r^2)$

$\displaystyle \frac{dA}{dr}=-32\pi(r^{-2})+4\pi(r)=0$

$\displaystyle -32\pi+4\pi(r^3)=0$

$\displaystyle 4r^3=32$

therefore, $\displaystyle r=2$

Is this correct?