Wait I'm having a problem working through this with a calculator. The amount of seed is reaching a limit of 33.33.
1 Day: 25 * .25 + 25 = 31.25
2 Days: 31.25 * .25 + 25 = 32.8125
3 Days: 32.8125 * .25 + 25 = 33.203
.....
LIMIT EXPLAINED:
Assume a quart has 100 seeds, so there will be 25 millet seeds in each quart of feeder.
The math for 1 day would be: (25 * .25 + 25) -> 25(.25 + 1)
The math for 2 days would be: (25 * .25 + 25) * .25 + 25 -> (.25^2 * 25 + .25 * 25) + 25 -> 25(.25 + .25^2 + 1)
The math for 3 days would be: 25(.25 + .25^2 + .25^3 + 1)
This infinitely repeating pattern S can be approximated with the following math (this is more accurate the more days millet is laid):
S = .25 + .25^2 + .25^3 + .25^4 ...
.25*S = .25^2 + .25^3 + .25^4 + .25^5 ...
Focusing on the term S, we can eliminate the infinitely repeating part of the equations by subtracting .25 * S from S. Subtracting both equations equals .25!
Solve for S: S - .25*S = .25 -> S(1 - .25) = .25 -> S(.75) = .25 -> S = .33
RETURN TO THE EQUATIONS ABOVE:
The equations above can be simplified to the following: 25(S + 1) = 25(1.33) = 33.25, which is why the amount of Millet reaches a limit of approximately 33.33.