Determine the minimum using the second derivative

[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!

 

[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.

 

[ranger1716]

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.

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.

 

[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.