# Minimizing Differentiation...can volume

• Aug 23rd 2010, 01:48 PM
redskies
Minimizing Differentiation...can volume
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:

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

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

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

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

$\displaystyle 4r^3=32$

therefore, $\displaystyle r=2$

Is this correct?
• Aug 23rd 2010, 02:02 PM
Quote:

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?

Yes. There are some typos for you to review which have been corrected above.
• Aug 23rd 2010, 02:04 PM