I am not sure if “holes” is a generic* joke.

Your response indicates that you are thinking about the package as necessarily being a rectangular prism. To ensure that we are referring to the same problem: The version about which I am curious is more general, and allows the cubical candies to be wrapped one side at a time.

To re-paste your original question,

>Given any number of cubical candies, what is the best way to minimize packaging?

The following image shows that the minimal surface area for 9 cubical candies is not achieved by using a rectangular prism:

http://i.stack.imgur.com/iz1Sv.jpg

That image comes from the following link, which also leads to your blog:

http://mathoverflow.net/questions/179371

MQ//Mathwater

*Despite the risk of over-explaining wordplay: This is, indeed, a “generic”/”genus” mathematical pun.

Have you ever returned to the integral version of this problem?

Frequently! Here is the solution I find most promising. I haven’t found an example to contradict it and I don’t have a proof that it works.

1. Take your number of candies.
2. Take the cube root of that number.
3. Find the factor of your number of candies that is closest to that cube root.
4. Find the quotient of the number of candies and the factor from [3].
5. Take the square root of that quotient.
6. Find the factor of your number of candies that is closest to that square root.
7. The third factor is trivial.
8. Use the smallest factor as your height.

See any obvious holes I’ve missed?

Have you ever returned to the integral version of this problem? I do not believe it is amenable to elementary means. (Perhaps it can be out-sourced to a combinatorialist, if you happen to know one.)

MQ//Mathwater

The algorithm already fails at n=7, which wants 1×2×4, with surface area 2(2+4+8)= 28, while a 2×2×2 box fits 7 (with a space) and area only 6(4) = 24.

If you only need to solve this for small numbers, then a simple computational approach is attractive: compute volume and area for all small boxes with integer side lengths, sort by area (increasing) and subkey volume (decreasing), and remove any from the list whose volume is less than an entry earlier on the list. This provides a list of the biggest volume you can contain for any given surface area (up to the maximum size box of interest).

The more general question is a Diophantine equation problem, which probably means it is difficult to solve analytically.

I love the way that this video expresses, both in the playful nature with which the candies dance around as well as in the brilliant song choice, a sophisticated passion for mathematics. It is clear that you care a great deal about your work.

As for purposeful practice, it is leveraged masterfully by the authors of the Philips Exeter Academy materials to create a curricular masterpiece. By embedding practice opportunities into problem solving situations, they have woven together a curriculum focused almost entirely on concept development, without any loss to skill development. Neither a word nor an inch of space is wasted in those texts. Working through them with the right modeling tools on hand is like having the world of mathematics unfold in space like a magical pop-up book. A world of pure imagination.

Works with 108 and 144 as well.