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The problem is made even more difficult by assuming that the bag is made out of a material like Paper or PET Film which can neither stretch nor Shear .

According to Anthony C. Robin , an approximate formula for the capacity of a sealed expanded bag is:

:V=w^3 \left (h/ \left (\pi w ight ) -0.142 \left (1-10^ \left (-h/w ight ) ight ) ight ),

where ''w'' is the width of the bag, ''h'' is the height, and ''V'' is the maximum volume.

A very rough approximation to the capacity of a bag that is open at one edge is:

:V=w^3 \left (h/ \left (\pi w ight ) -0.071 \left (1-10^ \left (-2h/w ight ) ight ) ight )

(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 ).


THE SQUARE TEABAG

In the special case where the bag is sealed on all edges and is square with unit sides, ''h'' = ''w'' = 1, and so the first formula estimates a volume for this of roughly:

:V= rac 1 {\pi} - 0.142 \cdot 0.9

or roughly 0.19. According to Andrew Kepert at the University Of Newcastle, Australia , the 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+.

In the article referred to above A C Robin also found a more complicated formula for the general paper bag. Whilst this is beyond the scope of a general work, it is of interest to note that for the tea bag case this formula gives 0.2017, unfortunately not within the bounds given by Kepert, but significantly nearer.


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