# Minimizing Surface Area

[SOLVED] Minimizing Surface Area

1. 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.

2. Homework Equations
V = $$\pi r^2 h$$
SA = $$2 \pi r^2 + 2 \pi r h$$

3. 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.
$$0 = 4 \pi r - 100 r^-2$$
$$r = \sqrt{\frac{100}{4 \pi}}$$
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.

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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.

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
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.