Minimizing Differentiation...can volume

**redskies** · August 23rd 2010, 01:48 PM

A company prodces cylindrical cans.The volume of each can must be
$16\pi(cm^3)$
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:
$2\pi(r)(h)+2\pi(r)^2$

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

Therefore :

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

Filling into Surface area:

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

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

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

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

$4r^3=32$

therefore, $r=2$

Is this correct?

**Archie Meade** · August 23rd 2010, 02:02 PM

Originally Posted by redskies

A company prodces cylindrical cans.The volume of each can must be
$16\pi(cm^3)$
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:
$2\pi(r)(h)+2\pi(r)^2$

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

Therefore :

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

Filling into Surface area:

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

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

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

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

$4r^3=32$

therefore, $r=2$

Is this correct?

Yes. There are some typos for you to review which have been corrected above.

**Archie Meade** · August 23rd 2010, 02:04 PM

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