Optimization - find the dimensions of a box as functions of volume - minimal surface area Had a basic calculus course exam today. This was one of the problems:
We have a rectangular box of a given volume V. Present the width, height, and length of the box as functions of V so that the box can be made with the least amount of materials. 
The box has a lid and we can assume the walls have no thickness.
I know this problem is somehow related to multivariable calculus. During the course we have learned the basics of multivariable functions: partial derivatives, directional derivatives, limits, local min/max values and such. How would you use these concepts to solve the task?
I'm pretty sure the box has to be cubical but can't prove it.
 A: The amount of material used to make the sides of the walls is proportial to the surface area of the box. If $l$ is the length, $w$ is the width, and $h$ is the height, then the volume is 
$$V = lwh$$
and the surface area (which we'd like to minimize) is
$$S = 2(lw+lh+wh)$$
We can reduce the minimization problem to minimizing a function of two variables if we write one of the three dimensions in terms of the volume and the other two. Let's arbitrarily choose height, so
$$h=\frac{V}{lw}$$
Now the surface area is a function of length and width 
$$S(l,w) = 2\left(lw + \frac{V}{w} + \frac{V}{l}\right) = 2\frac{l^2w^2 + Vl+Vw}{lw}$$
Given that you mentioned learning about local min/max values in some multivariable functions, you should be able to minimize $S(l,w)$ from here on. Feel free to ask more questions if you have any. Hope this helps!

By the way, your intuition that the box is cubical is correct. This is a generalization of the square being the rectangle that minimizes perimeter for a fixed area.
A: This is an easy application of the method of Lagrange multiplier. Suppose the box has width $a$, height $b$ and length $c$. You need to minimize the surface area
$$f(a,b,c)=2ab+2ac+2bc$$
subject to the constraint 
$$g(a,b,c)=abc=V.$$
We have
$$\nabla f=(2b+2c,2a+2c,2a+2b),\quad \nabla g=(bc,ac,ab)$$
and it is easy to see that a solution to $\nabla f=\lambda \nabla g$ is to take $a=b=c$ and $\lambda =\frac{4}{a}$. We then get that $a^3=V$, and so
$$a = V^{1/3},\quad b = V^{1/3},\quad c = V^{1/3}.$$
