You have two two-liter bottles,
one (marked W) containing one liter
of pure water and one (marked A)
containing one liter of pure alcohol.
The two things you can do are to
pour liquid from either bottle down
the drain or pour liquid from one
bottle into the other, but you must
always keep at least one liter of liquid
in bottle A.
What is the minimum
alcohol concentration that can be
achieved in bottle A? Assume that
there is no volume change on mixing
water and alcohol.
(In reply to
re(3): soln by Kenny M)
The proof is rigorous irrefutable logic.
There is only one way to dilute solution A: one adds a solution of a
lower concentration. The lower the concentration added and the more
added, the more the dilution occurs. The lowest concentration
that may be added is pure water and the maximum that may be added
is one liter.
Dilution also occurs maximally adding when A holds the smallest
volume of solution allowed - again, one liter. So, by adding infinitesimal portions of pure water, while draining infinitesimal
portions of solution from A, we meet all the optimization
conditions.
re(4): soln
