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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.

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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.

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  • $\begingroup$ Thanks! This was very helpful. $\endgroup$
    – user202261
    Dec 19, 2014 at 15:46
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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 $$a = V^{1/3},\quad b = V^{1/3},\quad c = V^{1/3}.$$

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  • $\begingroup$ 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. $\endgroup$
    – user202261
    Dec 19, 2014 at 15:47

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