How to get the correct solution of this simple equation in wxMaxima?

Question

I am trying to solve the following simple equation in wxMaxima:

21.3874=0.00202415/(d^3)

When I use solve(), I get 3 solutions (2 complex and 1 real). I know that d is real and positive (because it's a diameter) so I'm only interested in the third solution. It's given in the following form: d=((40483)^(1/3))/(10427748^(1/3)) and I tried saving it with map() function but it saved all 3 solutions and I don't know how to map only the last one so that I can use it for further calculations.

Anyway, after rewriting this solution and using float() function on it, the result is: 0.1571668 But it's incorrect since the result should be 0.0455717.

Interestingly, to_poly_solve() gives different solutions with the real one being: 0.2439467

I have 3 questions regarding this problem:

1. Why does wxMaxima give two incorrect solutions and how to get the right one ?
2. How can I tell the Maxima that I am looking only for real and positive solution of a given equation.
3. Is it possible to map selected solution so that it can be referenced in further calculations (for example to use float() on it) ?

Answer 1

Hmm, when I try it, I get the expected result (0.0455717).

(%i5) mysolutions: solve (21.3874 = 0.00202415/(d^3), d);

rat: replaced 21.3874 by 106937/5000 = 21.3874

rat: replaced -0.00202415 by -40483/20000000 = -0.00202415
                        1/3           1/3
           sqrt(3) 40483    %i - 40483
(%o5) [d = ------------------------------, 
                             1/3
                    20 427748
                             1/3           1/3             1/3
                sqrt(3) 40483    %i + 40483           40483
          d = - ------------------------------, d = ------------]
                                  1/3                        1/3
                         20 427748                  10 427748

In order to isolate just the third solution, note that solve returns a list, so you can just say mysolutions[3] (or whatever is the name of the variable to which the result of solve was assigned).

If you just want the numerical value, you can say rhs(mysolutions[3]) where rhs = "right-hand side". There is also lhs = "left-hand side".

However, note that it's convenient to just use the equation d = something as an argument to subst in order to substitute the value into some other expression, e.g.:

(%i6) preferred_solution: mysolutions[3];
                                   1/3
                              40483
(%o6)                   d = ------------
                                     1/3
                            10 427748
(%i7) float (preferred_solution);
(%o7)                d = 0.04557166543026844
(%i8) subst (preferred_solution, %pi*(d/2)^2);
                               2/3
                          40483    %pi
(%o8)                     -------------
                                    2/3
                          400 427748
(%i9) float (%);
(%o9)                 0.001631096598182001