# Maximum values of implicit functions in specific range

A simple implicit function

f = -0.000462963 x^2 - 0.1/Sqrt[0.0625 + (-3. + x)^2 + z^2]
- 0.1/Sqrt[0.0625 + (3. + x)^2 + z^2];
E0 = -0.0575;


and the corresponding contour plot

C0 = ContourPlot[f == E0, {x, -12, 12}, {z, -12, 12},
ContourStyle -> {{Black, Thickness[0.004]}}, AspectRatio -> 1,
ContourShading -> False, PlotPoints -> 100,
PerformanceGoal -> "Quality"] My target is to numerically obtain the maximum value of $z$ in the interval $-6 < x < 6$.

However when I use

max = NMaximize[{x, f == E0}, {{x, -6, 6}, z}];


the program compute a maximum value outside the desired interval. Why? Am I doing something wrong?

• What is Vxz in you Contourplot? – Julien Kluge Sep 28 '16 at 14:10
• @JulienKluge It's an error. See my edit. – Vaggelis_Z Sep 28 '16 at 14:12
• I think you mean E0 = - 0.0575 as well. – Marius Ladegård Meyer Sep 28 '16 at 14:23
• @MariusLadegårdMeyer You are right! – Vaggelis_Z Sep 28 '16 at 14:26

It was not nessecary to do it numerically. Solve with assumptions works.

Lets assume $x$ and $z$ to be $x,z\in\mathbb{R}$ and solving this:

sol = z /. Solve[f == E0, z, Reals];


This gives two solutions with a condition that $-11.1445 < x < 11.1445$. So we refine:

sol=Refine[sol,{11>x>-11}];


Lets differentiate it and take only the first or second function. (The second funtion is the one you want but both are symmetrical so it does not matter.)

dsol=D[sol[],x] Now we can search for solutions:

xsol=x/.Solve[dsol==0,x,Reals]


{{x->-11.1445},{x->-7.17004},{x->-3.16044},{x->0},{x->3.16044},{x->7.17004},{x->11.1445}}

So we see that $x\approx\pm 3.16044$ is your solution.

To get all pairs of solutions $(x,z)$ we map accordingly

solutionpoints=({#,sol[]/.x->#})&/@xsol


{{-11.1445,316.044},{-7.17004,0.305369},{-3.16044,2.61846},{0,0. +3.0104 I},{3.16044,2.61846},{7.17004,0.305369},{11.1445,316.044}}

which gives you your two solutions to be: $$(x_{1,2},z_{1,2})=(\pm3.16044,2.61846)$$ and displaying the result:

Show[Plot[sol,{x,-10,10}],ListPlot[solutionpoints]] EDIT: NMaximize would have also worked:

NMaximize[{z,f==E0,0<x<5},{x,z}]


{2.61846,{x->3.16044,z->2.61846}}

Which is the same solution.