@acer You are a real virtuoso in such things!

Obviously, the original expression is periodic with the period  2*Pi , because It contains only the functions of  sin(phi)  and  cos(phi). Therefore, it is sufficient to solve the equation, for example, in the range  0 .. 2*Pi  and continue periodically to any desired interval. The same applies to finding extremums (local or global).

More precisely: combine(ee) shows that the period is  Pi .

Edit.

@Rouben Rostamian   Convert your comment in an answer and you will get a lot of votes up for it. Mine be the first.

For example you can write:

s := Int(exp(-x^2)*cos(2*x*y), x = 0 .. infinity):
s=value(s);

L:=[[0.5,0],[1,0],[1,7],[7,6.5],[5,9],[12,12],[10,13],[12,17],[10,13],[12,17],[8,16],[8,18],[3,16],[5,23],[3,21],[0.5,27]]:
L1:=ListTools:-Reverse(map(t->[1-t[1],t[2]],L)):
plot([L[ ],L1[ ]], color="#DC2828", thickness=5, axes=none, scaling=constrained);
# Or
plot([L[ ],L1[ ]], color="#DC2828", thickness=5, filled, axes=none, scaling=constrained);

@Franzs  Just now I noticed this new thread that duplicates your previous question. Look there my comment.

@Franzs   Maple shows the specified color only those planes that are present when setting the body. Add inequality  z>=-40 :

with(PolyhedralSets);
P1 := PolyhedralSet({-x-20 <= 0, -x-20 <= 20-y, -x-20 <= -(1/2)*y, -x-20 <= -z, -x-20 <= -(x+y)*(1/2), -x-20 <= -(y+z)*(1/2), z>=-40});
Plot(P1, axes = normal, view = [-40 .. 40, -40 .. 40, -40 .. 40], orientation = [30, -30, -60], color = blue, thickness = 2);
Plot(P1, axes = normal, view = [-40 .. 40, -40 .. 40, -40 .. -40], orientation = [0, 0, 0], color = blue, thickness = 2);

Rotation of the initial figure around the z axis:

restart;
with(PolyhedralSets):
P1 := PolyhedralSet({-x-20 <= 0, -x-20 <= 20-y, -x-20 <= -(1/2)*y, -x-20 <= -z, -x-20 <= -(x+y)*(1/2), -x-20 <= -(y+z)*(1/2)}):
A:=Plot(P1, axes = normal, view = [-40 .. 40, -40 .. 40, -40 .. 40], orientation = [30, -30, -60], color = blue, thickness = 2):
f:=alpha->plottools:-rotate(A, alpha, [[0,0,0],[0,0,1]]):
plots:-animate(plots:-display, [f(alpha)], alpha=0..2*Pi, frames=90);

@asa12   You wrote: If change to F(exp(t)) = 2*t, is it possible?

No, because for  t=exp(1)  we have:

F(F(exp(t))) = F(2*t) = F(2*exp(1)) = F(exp(ln(2)+1) = 2*(ln(2)+1) <> 0

@Markiyan Hirnyk   Use  expand  command first:

y=0.0000125698-0.0000125698*cos(54*x);
a:=convert(%,fraction);
b:=content(rhs(a), cos(54*x));
c:=primpart(rhs(a), cos(54*x));
y=b*``(c);
convert(%, float, 6);
diff(expand(rhs(%)), x);
 

  @toandhsp The acer's method with  extrema  can also be used for this example, but one addition is needed. If the curve defining the relationship between the variables were smooth as in the initial example, then there would be no problems. But in this example there is the point  (x=4, y=3) lying on this curve in which the differentiability is broken. Therefore, this point must be checked separately for the presence of an extremum:

restart;
z:=x+I*y;
R:=evalc(3*abs(z-4-3*I)+4*abs(z-8-6*I) = 20);
is(eval(R, [x=4,y=3]));
is(eval(R, [x=8,y=6]));
a:=op(extrema(sqrt(x^2+y^2), {R}, {x,y}, 's'));  # The first candidate for an extremum
A:=[op(s[1])];
b:=simplify(eval(sqrt(x^2+y^2), [x=4,y=3]));  # The second candidate for an extremum
Min=min(a, b), Max=max(a, b);  # The results
plots:-display(plots:-implicitplot(R, x=0..9, y=0..8, color=blue, gridrefine=5), plottools:-disk([4,3], 0.1, color=red), plottools:-disk(rhs~(A), 0.1, color=red), view=[0..9,0..8]);  # Visualization 

@Markiyan Hirnyk  Sorry, but I do not understand your reaction. Maybe someone else can satisfy your curiosity.

@Markiyan Hirnyk   See wiki  https://en.wikipedia.org/wiki/Discrete_uniform_distribution

@John Fredsted  For  z=x+I*y  we have a continuous function  (x,y)->sqrt(x^2+y^2)  defined on the segment connecting the points  A(-3, 2)  and  B(3, 8) . In fact, this will be a function of one variable, because  [x,y]=(1-t)*[-3,2]+t*[3,8]  or  x=6*t-3, y=2+6*t  for t in [0,1] . If an extremum (min  or  max) is reached within a segment, then this will be a critical point for this function, that is, the derivative at this point is 0. But the extremum can be reached at the ends of the segment and the corresponding points do not need to be critical. In this example, the maximum is reached at the end of the segment for  t = 1 .

@John Fredsted  I think this is the simplest way to solve this problem. The difference in efficiency is not too big:

primesInRange := (a::posint,b::posint) -> select(isprime,[$(a..b)]):

t:=time():   
PrimesInRange(10^5,10^6):
time()-t;

t:=time():   
primesInRange(10^5,10^6):
time()-t;

                             2.750
                             5.375

 

@Carl Love  Should be  p<=b