I did the following. Subtract one side from the other to get expressions from equations, square each and sum. I want to use that sum of squares as my constraint equation.
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The somewhat lengthy equations may be found here. There also were variants here and here, but I used that first (I believe I did not get very good results with the others).
Anyway, to start out, define eqns = ... where ... denotes the list of equations (they were not in an explicit Mathematica List so one musty edit anyway). The rest proceeds as below.
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polys = Apply[Subtract, eqns, 1];
sos = polys.polys;
vars = Variables[polys];
Now numerically maximize q[2]. The setting below seem to work well but what I tried was by no means exhaustive. I did not even check whether the post process setting was useful.
{max, vals} =
NMaximize[{q[2], sos == 0}, vars, MaxIterations -> 2500,
Method -> {"DifferentialEvolution",
"PostProcess" -> "InteriorPoint"}]
(* Out[502]= {249.612839472, {q[1] -> 10.0776784587,
q[2] -> 249.612839472, q[3] -> 1.2374159599*10^-11,
q[4] -> 2.63182261366*10^-11, q[5] -> 3.92474408986*10^-11,
q[6] -> 5.59153450658*10^-11, q[7] -> 3.26994887506*10^-11,
q[8] -> -7.23283009989, q[9] -> -2.33832951438*10^-11,
q[10] -> -4.44339637424*10^-11, q[11] -> -4.86272772623*10^-11,
q[12] -> -5.045240705*10^-11, q[13] -> -4.72526447498*10^-11}} *)
Test quality of result by checking residuals.
polys /. vals
(* Out[503]= {0., 0.0000353733578184, 0.0000211878856078,
-4.78962640216*10^-7, -0.0000328136205994, -0.0000674022441487,
-0.000388164292971, -0.0000813888584545, -0.0000606146825248,
-0.0000387691273644, -0.0000236987141351, -0.0000166052409085} *)
To my surprise, nay, amazement, we have a very good approximate solution. Given the scaling of the inputs I was really not expecting that we could make all of the residuals that small.
Whether this is actually a useful result remains to be seen. The overdetermined system, with it's funny scaling, does not strike me as a great starting point for getting a good outcome in, say, an experimental science scenario.