The commercial macsyma can solve this, however it leaves the
answer in terms of rootof. However, it looks like the rootof expression
is a polynomial.
Following is the result:
[[b = 0, f = fmin], [b = 0, f = fmax], [b = bmin,
f = (( - bmin^2 * rb + bmin * (bmin + bmax) * rb - bmax * bmin * rb)/(a * bmax *
bmin))], [b = bmax,
f = ((bmax * (bmin + bmax) * rb - bmax * bmin * rb - bmax^2 * rb)/(a * bmax *
bmin))], [f = 0, b = 0],
[b = root_of(rf * rb^2 * b^4 + ( - 2 * rf * bmin - 2 * rf * bmax) * rb^2 * b^3
+ ((rf * bmin^2 + 4 * rf * bmax * bmin + rf * bmax^2) * rb^2 + (fmin + fmax) *
rf * a * bmax * bmin * rb) * b^2
+ (( - 2 * rf * bmax * bmin^2 - 2 * rf * bmax^2 * bmin) * rb^2
+ (( - fmin - fmax) * rf * a * bmax * bmin^2 + ( - fmin - fmax) * rf * a *
bmax^2 * bmin)
* rb + fmax * fmin * a^3 * bmax^2 * bmin^2) * b + rf * bmax^2 * bmin^2
* rb^2 + (fmin + fmax) * rf * a * bmax^2 * bmin^2 * rb + fmax * fmin * rf * a^2
* bmax^2 * bmin^2, b),
f = ((b * (bmin + bmax) * rb - bmax * bmin * rb - b^2 * rb)/(a * bmax * bmin))]]
David and Michele Holmgren wrote:
> Hi - Even if you try eliminate([f1,f2],[B]) or eliminate([f1,f2],[F]), you
> get some very ugly expressions which seem to defy simplfication or solution.
> Also eliminate([f1,f2],[B,F]) produces [0] as a result. So, this system may
> defy symbolic solution (unfortunately), and a numerical method seems the
> only recourse.
>
> Dave Holmgren
>
> > > I'm new to Maxima
> >
> > Thanks for your interest in Maxima!
> >
> > > f1: (Rb/Bmin) * B * (B - Bmin) * (1 - B/Bmax) - A * B * F = 0;
> > > f2: (Rf/Fmin) * F * (F - Fmin) * (1 - F/Fmax) - A * B * F = 0;
> > > res: solve ([f1, f2], [B, F]);
> > >
> > > Now if I input just that, and try to solve it symbolically,
> > > Maxima hangs.
> >
> > I think you have found a bug.
> >
> > Your system reduces to something of the form:
> >
> > q1: b2 * B^2 + b1 * B + b0 + bf * F = 0
> >
> > q2: f2 * F^2 + f1 * F + f0 + fb * B = 0
> >
> > and Maxima seems to run forever on that, too.
> >
> > I tried solving the system step by step.
> >
> > First, I solved for F in f1 and substituted. This leads to a quartic
> > mess.
> >
> > Then I tried solving for B in f1 and substituting into f2; Solve solves
> > the resulting equation easily. There are two non-trivial solutions in
> > each step, and you'll have to check which ones are meaningful.
> >
> > By the way, it can be helpful to declare some facts about your
> > parameters, e.g.
> >
> > assume(Bmax>B,B>Bmin,Bmin>0,
> > Fmax>F,F>Fmin,Fmin>0,
> > Rb>0,Rf>0,A>0)
> >
> > This will keep some radicals "clean", though it does *not* use these
> > facts in Solve (for example, you will get the B=0, F=0 solution).
> >
> > Good luck.
> >
> > -s
> >
> > _______________________________________________
> > Maxima mailing list
> > Maxima@www.math.utexas.edu
> > http://www.math.utexas.edu/mailman/listinfo/maxima
>
> _______________________________________________
> Maxima mailing list
> Maxima@www.math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima