gamestoper05 Posts: 1, Reputation: 1 New Member #1 Sep 16, 2007, 09:31 AM
Precalculus optimization
A container manufacturing company has been contracted to design and manufacture cylindrical cans for juice. The volume of each can is to be 0.170 liters. In order to minimize production costs, the company wishes to design a can that requires the smallest amount of material possible. What should the dimensions of the can be?

I know I need to know the surface area of the can. I know it's a lot to ask, but can someone please walk me through this? Its due tomorrow!:confused:
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Sep 16, 2007, 10:09 AM
The surface area of a cylinder with top and bottom is given by:

$S=2{\pi}rh+2{\pi}r^{2}$... [1]

Volume is: $V={\pi}r^{2}h$

$0.17={\pi}r^{2}h$... [2]

I will tell you that the minimum surface area occurs when the height is twice the radius. h=2r

Then your volume is $\frac{17}{100}={\pi}r^{2}(2r)$

Solve for r and you have the radius which will give minimum surface area. Twice that will be the height.

Easier than you thought, huh?

The proof that the minimum surface area occurs when h=2r can be done without calculus.

Commonly done this way:

Since $h=\frac{V}{{\pi}r^{2}}$, we have

$S=2{\pi}r^{2}+2{\pi}r(\frac{V}{{\pi}r^{2}})$

$S=2{\pi}r^{2}+2(\frac{V}{r})$

We have equality when $2{\pi}r^{2}=\frac{V}{r}$ (this uses the arithmetic-geometric mean which I will not bother delving into).

$2{\pi}r^{3}=V$

But $V={\pi}r^{2}h$

So, we have $2{\pi}r^{3}={\pi}r^{2}h$

$h=2r$
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