Oops. I said:
...substituting into f2; Solve solves the resulting equation
easily...
Which is true. Solve *did* solve the equation easily. Unfortunately,
it did not solve it completely -- there is still an F on the right-hand
side. This brings us back to the messy quartic, which did not magically
cancel away.... And of course the general symbolic solution of a
quartic is a huge mess. Maxima can calculate it (trust me, I did it),
but the result is enormous and presumably useless. One way to approach
it is to just take the resulting quartic and run radcan over it for a
few days, in the hopes that something will simplify. I wouldn't be too
optimistic about that.
Another approach is to approximate. For example, for A << 1, I get the
following solution (among others):
A Bmax Bmin (Fmin + Fmax)
B = Bmax - -------------------------
(Bmax - Bmin) Rb
I applied the approximation after the substitution, and before the
second Solve.
But I have been quick and dirty and sloppy in deriving this, so it is
perfectly possible I've blundered again. It does come out to within
0.1% of one of the numeric solutions -- is that good enough?
-s