# How to define and plot a maximum function?

Define a function $g$ as follows:

f = 2/((x1 + Sqrt x2)^2 + (y1 + Sqrt y2)^2)
X = vx1^2*D[D[f, x1], x1] + vx2^2*D[D[f, x2], x2] + 2 vx1*vx2*D[D[f, x1], x2]
Y= vx1*D[f, x1] + vx2*D[f, x2]
g=X+Y^2


The function $g$ is clearly a function in 6 variables ($vx1,vx2,x1,x2,y1,y2$).

Now I want to define a function in two variables $max$ as the maximum of $g$ for $x1,x2,y1,y2$ which vary in the following hypersurface

H = ImplicitRegion[{3 (-x1^2 - y1^2 + x2^2 + y2^2) +
2 Sqrt (x1*x2 + y1*y2) == 0}, {x1, y1, x2, y2}]


So I want to (but I don't know how to) define the function $max:\mathbb{R}^2\rightarrow \mathbb{R}$ as follows:

$$max(vx1,vx2)=\max\limits_{(x1,x2,y1,y2)\in H}g(vx1,vx2,x1,x2,y1,y2)$$

All I could think of is the command

FindMaximum[{g, {x1, y1, x2, y2} ∈ H}, {x1, y1, x2, y2}]


but this doesn't define a function, it just gives the maximum of $g$ for specific values of $vx1,vx2$.

How can I define the function $max$? It's important to me to define it as a function because I need to

• derive it in its smooth points

• plot it

• define other functions which involve $max$ in their definition

How can I do it?

EDIT: I followed the suggestions of user Corey979 and defined the function max[vx1,vx2], but with FindMaximum substitued by MaxValue, since I'm interested in the maximum value (as I thought was clear in my definition). But then I'm not able to do any operation with the function max:

Any of these operations

D[max[vx1, vx2], vx1]

Plot3D[max[vx1, vx2], {vx1, -10, 10}, {vx2, -10, 10}]

FindMaximum[{max[vx1, vx2]}, {vx1, vx2}]


Will require an extremely long computational time and will give no output. Which is strange, since with my old code

f = 2/((x1 + Sqrt x2)^2 + (y1 + Sqrt y2)^2)
X = vx1^2*D[D[f, x1], x1] + vx2^2*D[D[f, x2], x2] + 2 vx1*vx2*D[D[f, x1], x2]
Y= vx1*D[f, x1] + vx2*D[f, x2]
g=X+Y^2
H = ImplicitRegion[{3 (-x1^2 - y1^2 + x2^2 + y2^2) +
2 Sqrt (x1*x2 + y1*y2) == 0}, {x1, y1, x2, y2}]
FindMaximum[{g, {x1, y1, x2, y2} ∈ H}, {vx1,vx2,x1, y1, x2, y2}]


I was able to get the answer

{2.39111, {vx1 -> 1.55608, vx2 -> 1.96316, x1 -> 0.810245, y1 -> -0.236216, x2 -> 1.34919, y2 -> -0.463331}}


Is there a better way to define max in such a way that it will be possible to perform operations with it?

• What are some typical values of vx1 and vx2? Jan 17, 2017 at 10:38
• Let's just assume that the domain of $h$ is $\mathbb{R}^2$ Jan 17, 2017 at 12:21
• The underscore Blank (_) is used when defining functions, like f[x_]:=x^2; then you call it normally with f[x]. You can't have them in expressions like D[max[vx1_, vx2_], vx1] or FindMaximum[{max[vx1_, vx2_]}, {vx1, vx2}]. That's a simple syntax error. Jan 17, 2017 at 12:22
• Also, you should read the docs of FindMaximum as you misunderstand its output - with my def of max, FindMaximum[{max[vx1, vx2]}, {vx1, vx2}] makes no sense. Jan 17, 2017 at 12:28
• I corrected the $_$ error. As I wrote in the definition of $h$, $h$ is defined as the maximum value of g. should I substitute FindMaximum with MaxValue in your code? If yes, then I'm not able to do any computation with max function, because the computational time is too long Jan 17, 2017 at 12:41

f[x1_, x2_, y1_, y2_] :=
2/((x1 + Sqrt x2)^2 + (y1 + Sqrt y2)^2)

X[vx1_, vx2_, x1_, x2_, y1_, y2_] :=
vx1^2*D[D[f[x1, x2, y1, y2], x1], x1] +
vx2^2*D[D[f[x1, x2, y1, y2], x2], x2] +
2 vx1*vx2*D[D[f[x1, x2, y1, y2], x1], x2]

Y[vx1_, vx2_, x1_, x2_, y1_, y2_] :=
vx1*D[f[x1, x2, y1, y2], x1] + vx2*D[f[x1, x2, y1, y2], x2]

g[vx1_, vx2_, x1_, x2_, y1_, y2_] :=
X[vx1, vx2, x1, x2, y1, y2] + Y[vx1, vx2, x1, x2, y1, y2]^2

H = ImplicitRegion[{3 (-x1^2 - y1^2 + x2^2 + y2^2) + 2 Sqrt (x1*x2 + y1*y2) == 0}, {x1, y1, x2, y2}]


Define the max as a function:

max[vx1_, vx2_] :=
FindMaximum[{g[vx1, vx2, x1, x2, y1,
y2], {x1, y1, x2, y2} ∈ H}, {x1, y1, x2, y2}]


Then

m = max[1000., 1000.] (The value of the maximum is a numerical zero; you might want to incorporate Chop in max.) Verify that the solution is in H:

m[[2, All, 2]] ∈ H


True

• thank you for your answer, but doing as you said I then encountered some problems trying to do operations with function max. I edited my question because it took too long to explain the problems in comments, I hope it's all right. Jan 17, 2017 at 12:18