In geometry, the paper bag problem or teabag problem is to calculate the maximum possible inflated volume of an initially flat sealed rectangular bag which has the same shape as a cushion or pillow, made out of two pieces of material which can bend but not stretch.
According to Anthony C. Robin, an approximate formula for the capacity of a sealed expanded bag is:
V=w3\left(h/\left(\piw\right)-0.142\left(1-10\left(-h/w\right)\right)\right),
where w is the width of the bag (the shorter dimension), h is the height (the longer dimension), and V is the maximum volume. The approximation ignores the crimping round the equator of the bag.
A very rough approximation to the capacity of a bag that is open at one edge is:
V=w3\left(h/\left(\piw\right)-0.071\left(1-10\left(-2h/w\right)\right)\right)
(This latter formula assumes that the corners at the bottom of the bag are linked by a single edge, and that the base of the bag is not a more complex shape such as a lens).
For the special case where the bag is sealed on all edges and is square with unit sides, h = w = 1, the first formula estimates a volume of roughly
| V= | 1 |
| \pi |
-0.142 ⋅ 0.9
or roughly 0.19. According to Andrew Kepert at the University of Newcastle, Australia, an upper bound for this version of the teabag problem is 0.217+, and he has made a construction that appears to give a volume of 0.2055+.
Robin also found a more complicated formula for the general paper bag, which gives 0.2017, below the bounds given by Kepert (i.e., 0.2055+ ≤ maximum volume ≤ 0.217+).