Homework Help: Determine the minimum using the second derivative

1. Sep 24, 2006

ranger1716

I was wondering if someone could show me where to go next in this problem.

I need to determine the minimum length, width and height that a 1 cubic foot box can have. This box does not have a top. I know that I need to minimize the area, but I'm not sure if I'm going about this correctly. So far I have that A(l,w)=lw+2(1/l)+2(1/w). I made the substitution from three to two variables because V=lwh therefore h=1/lw.

I'm assuming that I need to take the derivative of the area equation, and then determine the minimum using the second derivative. Is this the correct procedure?

Thanks!

2. Sep 24, 2006

0rthodontist

You need to take the derivative of the area equation and find where it's 0. Then you can check with the second derivative to ensure that that point is a minimum.

3. Sep 24, 2006

ranger1716

That's what I was thinking, however I don't know how to do this with two variables. I'm in Calc 3 but we haven't yet done partial derivatives. Everything that I have seen (including using Maple) needs to use partial derivatives to solve the problem. I've basically attempted to take the derivative using Maple, and got a partial derivative. I don't really know where to continue on to.

Last edited: Sep 24, 2006
4. Sep 24, 2006

0rthodontist

Oh, right, you do need partial derivatives, using the Hessian. On the other hand you could make the assumption l = w, which you could prove--start by saying that if l > w (without loss of generality) then there is a cheaper box with the same volume and height where l = w.