So I was in my cousin’s office this weekend and I saw a 5 gallon water bottle that had been emptied of water and filled with spare change. My first thought was that it must be terribly heavy, and when I tried to lift it I discovered I was correct – it’s extremely heavy. I was able to tip it to get my hands under it, but I was unable to lift it. Perhaps with help from another person I could even carry it around.
The bottle is 18.93 liters, give or take, which is .667 cubic feet. Combined with this page I have what I consider to be a lower bound for the value of the bottle: $328, if it’s full of pennies and not packed optimally. I could probably compute an upper bound as well, assuming it’s made of the highest value-per-volume coin (I think that’s going to be dimes, but I’m not sure).
If I were able to obtain an accurate weight for the full jug (and a reasonably precise estimate for the mass of the empty jug), how closely could I estimate the value of all of the coins inside? I’d like to think that simple linear algebra would do the trick, but I can only think of three equations – one for total mass of the coins, one for volume, and one for total value – and I know I’ve got at least four unknowns (number of each type of coin).
.25q + .10d + .05n + .01p = $(some large number)
Mqq + Mdd + Mnn + Mp * p = (some large number of kilograms), where M(x) is the mass, in kilograms, of coin x.
Vqq + Vd * d + Vn * n + Vp * p = .667 cubic feet, where V(x) is the volume, in cubic feet, of coin x.
M(x) and V(x) are almost certainly public information, but that still leaves me with three equations and five unknowns, even if I do know the mass to within a few grams. I think my only hope is to find an equation governing the frequency with which each coin is circulated relative to the others. Assuming that’s possible, could one estimate the value of the coins inside without knowing the mass of the full jug at all?