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I've the following code:

NSolve[{1/
    Sqrt[(1 - ((2*Cx*56000^2*(Cx + C3)))*4*Pi^2*60^2)^2 + 
      4*Pi^2*60^2*(((4*Cx*56000)) - (Cx^2*C3*56000^3)*4*
           Pi^2*60^2)^2] == 
   1/Sqrt[2], (200704000000000000000000000000 Cx^2 60^2 \[Pi]^2 (1 - 
         11289600000000000000000000000000 C3 Cx \[Pi]^2)^2 + (1 - 
        90316800000000000000000000000000 Cx (C3 + 
           Cx) \[Pi]^2)^2)/(7225344000000000000000000000000000000 \
Cx^2 \[Pi]^2 (1 - 
         112896000000000000000000000000000000 C3 Cx \[Pi]^2)^2 + (1 - 
        903168000000000000000000000000000000 Cx (C3 + 
           Cx) \[Pi]^2)^2) == 10^((s)/(10)), 
  Cx > 0 && C3 > 0 && 0 < C3 < 3 Cx}, {Cx, C3}]

Question: I need to optimize $s$. I now that $s$ is between $-60\le s\le-59$. But how can I aprove my code that is automatically solves the equation with the value for $s$ that is closesed to $-60$?

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  • $\begingroup$ Why not add the constraint for s in your NSolve[]? $\endgroup$
    – J. M.'s slightly less busy ♦
    Mar 27, 2018 at 8:49
  • $\begingroup$ @J.M. When I substitute $-59$ or $-59.99$ for $s$ I get 'no solution'. So it would be better if it choose the value for $s$ for itself closesed to $-60$ $\endgroup$
    – Looper
    Mar 27, 2018 at 8:53

1 Answer 1

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If we name the first equation eq1, ContourPlot shows the possible solutions {Cx,C3}

eq1 = 1/Sqrt[(1 - ((2*Cx*56000^2*(Cx + C3)))*4*Pi^2*60^2)^2 + 
 4*Pi^2*60^2*(((4*Cx*56000)) - (Cx^2*C3*56000^3)*4*
      Pi^2*60^2)^2] == 1/Sqrt[2]   
ContourPlot[ Evaluate[eq1 ] , {Cx, 0, 3 10^-8}, {C3, 0, 10^-7},FrameLabel -> {Cx, C3}, PlotRange -> {0, Automatic}]

enter image description here

Now you can substitute the solution 

ergC3 = Solve[eq1, C3](* solution C3[Cx]*)

into the second equation and solve for s[Cx]

eq2 = (200704000000000000000000000000 Cx^2 60^2 \[Pi]^2 (1 - 
      11289600000000000000000000000000 C3 Cx \[Pi]^2)^2 + (1 - 
     90316800000000000000000000000000 Cx (C3 + 
        Cx) \[Pi]^2)^2)/(7225344000000000000000000000000000000 \Cx^2 \[Pi]^2 (1 -112896000000000000000000000000000000 C3 Cx \[Pi]^2)^2 + (1 \- 903168000000000000000000000000000000 Cx (C3 + Cx) \[Pi]^2)^2) ==10^((s)/(10));

ergs = Solve[eq2/.ergC3, s, Reals] [[1]];ergs = Solve[eq2, s, Reals]   /.ergC3;
Plot[{(*C3,*)s /. ergs}, {Cx, 0, 3 10^-8}, PlotRange -> {-150, 0},AxesLabel -> {Cx, s}]

enter image description here

The plot shows s~-120!

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