Optimizing Can Dimensions for Minimizing Material Usage

[Sheneron]

[SOLVED] Minimizing Surface Area

Homework Statement

A can is to be manufactured in the shape of a circular cylinder with volume = 50.
Find the dimensions of a can that would minimize the amount of material needed to make the can.

Homework Equations

V = $$\pi r^2 h$$
SA = $$2 \pi r^2 + 2 \pi r h$$

The Attempt at a Solution

I have never done a problem like this so I am unsure how to do it, but here is my attempt.

With the volume equation I solved for h. $$h = \frac{50}{\pi r^2}$$
I plugged this value for h into the Surface area equation. $$SA = 2 \pi r^2 + 2 \pi r \frac{50}{\pi r^2}$$

which = $$2 \pi r^2 + \frac{100}{r}$$

I then took the derivative of that and set it equal to 0.
$$0 = 4 \pi r - 100 r^-2$$
$$r = \sqrt[3]{\frac{100}{4 \pi}}$$
r = 1.996

Then I plugged that back into the volume equation to solve for h and got h= 3.99.
Could someone tell if this is right and if not where I went wrong. Thanks.

 

[Snazzy]

Looks good to me. Check your result by plugging in a number smaller than r and a number bigger than r into the derivative. If a number smaller than r makes the derivative negative and a number larger than r makes the derivative positive, then r would be a minimum.

 

[Sheneron]

So is that how I know that its a minimum instead of a maximum. I'm still a little confused about that. What would I do if I wanted to solve this problem for a maximum?

 

[HallsofIvy]

There is no maximum.

 

[Sheneron]

Does it not have one? Even though the volume is set to 50 there is no max Surface area?

 

[Snazzy]

If you keep increasing r, the surface area just gets bigger and bigger without bound. The height gets smaller, but there is a separate $$2\pi r^2$$ term in the surface area calculation. That's why there is no maximum surface area.

 

[Sheneron]

I understand I think. h could get infinitely small which would make the surface area infinitely large.