# Maximization/minimization/whatever

• Feb 25th 2009, 02:42 PM
Maximization/minimization/whatever
"The volume of a cylindrical tin can with a top and a bottom is to be 16pi cubic inches. If a minimum amount of tin is to be used to construct the can, what must be the height, in inches, of the can?"

I know I need to take the derivative of the volume of a cylinder.
After that, though, I'm lost.
• Feb 25th 2009, 03:27 PM
Jester
Quote:

"The volume of a cylindrical tin can with a top and a bottom is to be 16pi cubic inches. If a minimum amount of tin is to be used to construct the can, what must be the height, in inches, of the can?"

I know I need to take the derivative of the volume of a cylinder.
After that, though, I'm lost.

Let $r$ be the radius and $h$ the height of the can. The volume is fixed so

$V = \pi r^2 h = 16\pi$

The surface area is

$A = 2 \pi r^2 + 2 \pi r h$

From the volume formula $h = \frac{16}{r^2}$ so the area is

$A = 2 \pi r^2 + \frac{32 \pi}{r}$

Now use calculus to find the r that minimizes A. Once you have that you can use the volume formula to find h.
• Feb 25th 2009, 03:41 PM
Quote:

Originally Posted by danny arrigo
Let $r$ be the radius and $h$ the height of the can. The volume is fixed so

$V = \pi r^2 h = 16\pi$

The surface area is

$A = 2 \pi r^2 + 2 \pi r h$

From the volume formula $h = \frac{16}{r^2}$ so the area is

$A = 2 \pi r^2 + \frac{32 \pi}{r}$

Now use calculus to find the r that minimizes A. Once you have that you can use the volume formula to find h.

That's where the problem comes in. I don't know how to do that.