## Minimize the Surface of a Cubuid

The problem I discuss in this blog post is from the nice page by Dan Meyer. The blog is interesting anyway because of the way Dan thinks about teaching. The concept „less helpful“ says it all. I was sometimes asked by a newbie teacher how to plan a course. My answer was: „Make them work!“. And with „work“ I am not talking about stupid fact learning. It is all about questions, discoveries, and the satisfaction to have found an answer, even it is not the best one. They will appreciate the solution only after having appreciated the problem.

Sorry about this digression. The problem is to minimize the surface

\(S = 2(ab+bc+cd)\)

given the volume

\(V = abc.\)

With no further restrictions the answer is …

That is one of the points where Dan would allow guesswork. Even better, he would allow you to discover the problem itself from real world situations. And to see, why it is a problem at all.

You can surf the answer, or find it using Lagrange’s method or any other method involving Calculus. The result is no surprise.

\(a=b=c=V^{1/3}.\)

If students do this I often have to ask the questions they do not ask. E.g., why is the solution of the Lagrange equations a minimum? Couldn’t it be a maximum? What happens if a, b, or c tend to 0? Can you find a proof without derivatives?

But the cited page does pose the constraint that a,b,c should be integer. This changes the problem and puts it to a higher level. In fact, I am quite sure that there is no easier algorithm than trying all triplets of divisors of V. There are some shortcuts, for sure, and to find them is a nice job. For a start, try simple numbers like 24.

The twist on the page is something deeper, however. Dan asks the readers for easy algorithms, and then puts up a contest to break these algorithms. This leads to endless trials and errors, discovering nice and useful math along the way. I like this very much.

Let me give you an example. The most obvious candidate for a solution is to take divisors close the cube root of V, unless the remaining divisor is a large prime. You need to go a long way to find counterexamples to this. I had to write a program to find one, 5850. The minimal solution does not contain the divisors closest to the cube root, and none is a prime. But read the comments on that page for yourself.

How this all translates into teaching at a university, I do not know. But I know that it should.