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This is my code below. I'm trying to find the min and max values of z0value1 in the range of cmin<c<cmaxthat satisfies the equation and the inequality inside the Max,MinValue functions. I tried plotting the graph for the equation v0==... so from seeing that, I'm expecting z0value1 to be in the range of around [0.000335,0.000346]. How do I make my MinValue, MaxValue functions work properly?

Update

v0 = 5*10^-9;
{avalue, bvalue} = {0.00123, 0.00109};
{cmin, cmax} = {0.000814, 0.00109};

z0min = 
  MinValue[
    {z0value1, 
     v0 == 2/3 π avalue^2 bvalue + 2/3 π avalue^2 c - 
       (π avalue^2 (2 c^3 - 3 c^2 z0value1 + z0value1^3))/(3 c^2) 
     && cmin < c < cmax}, 
    c];

z0max = 
  MaxValue[
    {z0value1, 
     v0 == 2/3 π avalue^2 bvalue + 2/3 π avalue^2 c - 
       (π avalue^2 (2 c^3 - 3 c^2 z0value1 + z0value1^3))/(3 c^2) 
     && cmin < c < cmax}, 
    c];

{z0min, z0max}
{MinValue[
   {z0value1, 
    1/200000000 == 
      3.45379*10^-9 + 3.16861*10^-6 c - 
        (1.58431*10^-6 (2 c^3 - 3 c^2 z0value1 + z0value1^3))/c^2 
     && 0.000814 < c < 0.00109},
    c], 
 MaxValue[
   {z0value1, 
    1/200000000 == 
      3.45379*10^-9 + 3.16861*10^-6 c - 
        (1.58431*10^-6 (2 c^3 - 3 c^2 z0value1 + z0value1^3))/c^2 
     && 0.000814 < c < 0.00109}, 
    c]}
ContourPlot[
     v0 == 2/3 π avalue^2 bvalue + 
       2/3 π avalue^2 c - (π avalue^2 (2 c^3 - 3 c^2 z0value1 + 
          z0value1^3))/(3 c^2), {c, cmin, cmax}, {z0value1, 0.00033, 
      0.00035}, FrameLabel -> {c, z0value1}]

enter image description here

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    $\begingroup$ What do you mean by "How do I make my MinValue, MaxValue functions work properly? ". MinValue and MaxValue are built-in function, not your functions. You don't have any function definitions in your posted code. $\endgroup$
    – m_goldberg
    Jan 13, 2017 at 6:37
  • $\begingroup$ My bad. bad wording. I meant how do I get the output that I want by using those functions(or anything else) because they don't give me any output.. $\endgroup$
    – Jun
    Jan 13, 2017 at 6:40
  • $\begingroup$ Ok yes good point, I editted it. $\endgroup$
    – Jun
    Jan 13, 2017 at 7:03

1 Answer 1

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The objective function as given has no dependence on c. The code will work if z0value1 is made into an independent variable.

v0 = 5*10^-9;
{avalue, bvalue} = {0.00123, 0.00109};
{cmin, cmax} = {0.000814, 0.00109};

z0min = MinValue[{z0value1, 
    v0 == 2/3 Pi avalue^2 bvalue + 
       2/3 Pi avalue^2 c - (Pi avalue^2 (2 c^3 - 
            3 c^2 z0value1 + z0value1^3))/(3 c^2) && 
     cmin < c < cmax}, {c, z0value1}];

z0max = MaxValue[{z0value1, 
    v0 == 2/3 Pi avalue^2 bvalue + 
       2/3 Pi avalue^2 c - (Pi avalue^2 (2 c^3 - 
            3 c^2 z0value1 + z0value1^3))/(3 c^2) && 
     cmin < c < cmax}, {c, z0value1}];

{z0min, z0max}

(* {-0.00203336, 0.0016974} *)

Check the corresponding plot looks plausible

ContourPlot[
 v0 == 2/3 Pi avalue^2 bvalue + 
   2/3 Pi avalue^2 c - (Pi avalue^2 (2 c^3 - 3 c^2 z0value1 + 
        z0value1^3))/(3 c^2), {c, cmin, cmax}, {z0value1, z0min, 
  z0max}, FrameLabel -> {c, z0value1}, 
 Epilog -> {Red, PointSize[Medium], 
   Point[{{cmax, z0min}, {cmax, z0max}}]}]

Mathematica graphics

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