# 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.

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.

• Thanks! This was very helpful. Dec 19, 2014 at 15:46

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 $$a = V^{1/3},\quad b = V^{1/3},\quad c = V^{1/3}.$$

• Thanks! We didn't have the time to go through the Lagrange method during the course but I'll be sure to look it up now. Dec 19, 2014 at 15:47