Find the Largest Cube Volume on n x n Paper

[recon]

If I were to draw a complete, continuous net of a cube on a piece of paper measuring n by n, how can I proceed so that the resulting cube has the largest possible volume achievable from that paper size?

I know that the most obvious solution (at least to me) is to draw the net along a diagonal of the square piece of paper (meaning that some lines forming the net will be parallel to the diagonal, and others perpendicular). However, I've been in the world long enough to know that common sense does not usually dictate the right answer.

 

[Galileo]

If I calculated correctly, drawing the net along the diagonal will give squares with length:

$$\left(\frac{\sqrt 2}{4+\sqrt 3}\right) n \approx 0.247 n$$

which means you're better off taking the net horizontally or vertically, since then the length will be n/4=0.25n.

 

[DaveC426913]

Are you stuck with that particular net? There are other ways to build a cube from a flat piece of paper that might be more efficient. One way lays the squares out as mroe of a 'z' than a 't'.


P.S. I am an old, old man - in my thirtes at least. When I was young, there was no such thing as a 'net'. I was helping edit an elementary schoolbook last year, and the editor and I both came upon this word we had never encountered before. How it's in elementary schools? When did it get introduced?

 

[NateTG]

If I calculated correctly, drawing the net along the diagonal will give squares with length:

$$\left(\frac{\sqrt 2}{4+\sqrt 3}\right) n \approx 0.247 n$$

which means you're better off taking the net horizontally or vertically, since then the length will be n/4=0.25n.

If you don't need the net to be continuous, the maximum side length is going to be
$$\frac{n}{\sqrt{6}} \approx .4 n$$
It's possible to get a side length of
$$\frac{n}{2\sqrt{2}} \approx .35n$$
if the net forms an 'X' diagonally across the square piece of paper, but it involves splitting one of the faces into four pieces. That's likely to be optimal, but I don't have a proof handy. Martin Gardner discusses this problem in one of his books.