Smallest square to wrap a cylinder

Suppose you need to gift-wrap a cylinder (e.g., a can of tennis balls, or a large candle) of height $h$ and radius $r$. Here wrap is the natural sense of covering the surface area of the cylinder completely, without cutting the square, creasing however needed. What is the smallest square that suffices for a given $h$ and $r$? For example, a rectangle of dimensions $(h+2r) \times (2 \pi r)$ suffices for how one might wrap a can of tennis balls or a stout candle:  In this $h=3$ and $r=1$ case, the rectangle has dimensions $5 \times 6.28$, and so a square of side $2 \pi$ suffices. But is that optimal?

Merry Christmas!

• @Alexander E.'s edit: I did mean square, for the smallest-area rectangle, can, I believe, approach the surface area of the cylinder, by using a long, thin rectangle. So I took the liberty of rolling back. – Joseph O'Rourke Dec 25 '12 at 21:24
• @Joseph: sorry, now I understand. I removed my edit. – Alexandre Eremenko Dec 25 '12 at 21:29
• @Alexandre: No problem; and sorry for misspelling your name! – Joseph O'Rourke Dec 25 '12 at 21:46
• If the paper has pictures of, say, reindeer, do we need to have them come out entire on either the top or bottom or cylindrical part, or can we bend them? – Will Jagy Dec 25 '12 at 22:16
• @Will: That would greatly complicate the question, to require matching patterns! Let's pursue that after solving the easier question I posed. – Joseph O'Rourke Dec 25 '12 at 22:20

You can wrap the cylinder with a square whose diagonal has length $2(h + 2r)$: put the cylinder in the center of the square and fold the corners up to meet at the top of the cylinder. When $h/r$ is less than $\pi - 2$, this method uses less paper than the OP's method. If $h/r$ is very small (if you're wrapping a CD, for example), it uses just a little over $2/\pi$ as much paper as the OP's method.

I haven't done a comparison with Gerhard Paseman's method...

Perhaps Joseph can fill in this description with a picture.

In square with vertices A, B, C, and D in clockwise order, draw a line from A to a point P on BC. If the length of AP is longer than the circumference R of the inscribed circle of ABP, then a cylinder with circumference R and height which I leave you to determine can be wrapped by ABCD. (Image added by J.O'Rourke)

Gerhard "Wishing You A Happy New Year" Paseman, 2012.12.25

• To answer Will's earlier question, you can bend it as desired. Alternatively, trace the pattern (say, something part of which has been provided by Joseph above) on the wrapping paper first, and then trace the square around the pattern. You can place the reindeer parts as desired. Gerhard "And A Merry Boxing Day" Paseman, 2012.12.25 – Gerhard Paseman Dec 26 '12 at 5:27
• Also, it is nice to have the circle of circumference AP also be interior to ABP, but it isn't necessary. The circle can be broken into parts, all of which are tangent to AP, but the assembled result does not have to be inscribable in ABP. It may be required that the circle diameter perpendicular to AP be inside ABCD, however. Gerhard "Ask Me About Reusing Paper" Paseman, 2012.12.25 – Gerhard Paseman Dec 26 '12 at 5:34

This is in response to Vectornaut's question about wrapping with a thin rectangle. In Geometric Folding Algorithms: Linkages, Origami, Polyhedra, it is argued (Theorem 15.2.1, p.236) that any polyhedron can be covered with a thin strip with arbitrarily small surface area beyond that of the polyhedron. A polyhedral approximation to a cylinder then yields the claim. This is Figure 52.2 (p.234), which gives some idea of switchback turns of the strip used in the argument: Here is a link to the original 1999 paper by Demaine, Demaine, and Mitchell, "Folding Flat Silhouettes and Wrapping Polyhedral Packages: New Results in Computational Origami": link.