@one man 

With your prefered method (where x,y are supposed to be Re and Im for z) ==> complex values for x,y.
Are you really satisfied with this? Note that Maple uses complex numbers by default.

@acer 

Is seems that you found a mistake in the code, but this would solve only the first bug (and partially the third, because this way at least `infinity` is replaced by an unevaluated `minimize`).

But Maple should be able to compute all three examples, because it can find the critical points:
f:=(x+y)^2+cos(y):
s:=solve([diff(f,x),diff(f,y)],[x,y], allsolutions);
      s := [[x = -Pi*_Z1~, y = Pi*_Z1~]]
eval(f,s[]);
      (-1)^_Z1~

The OP probably does not realize that  w=f(r), z=g(r)  is a parametric representation of a curve in the (w,z) space and its graph it's what he wants.

Example:   w=r+sin(6*r), z=r+sin(7*r),   r in [0,2*Pi];

plot([r+sin(6*r), r+sin(7*r) ,r=0..2*Pi], labels=["w","z"]);

@Markiyan Hirnyk 

Because the polynomials are very simple. But try this for

n^17-n^12-n^5+7
and
a^17-85*a^16+3400*a^15-85000*a^14+1487500*a^13-19337501*a^12+193375060*a^11-1519376650*a^10+9496121250*a^9-47480778125*a^8+189924350000*a^7-604311312500*a^6+1510804062499*a^5-2905466796850*a^4+4150820312250*a^3-4151035155000*a^2+2594580075000*a-763183590618

Here it is easier by hand!

@Markiyan Hirnyk 

If in Mathematica one wants to retrive Kitonum's solutions, a similar by hand substitution will be needed.

@Markiyan Hirnyk 

You are right. I have corrected the second example, which was answered correctly by Maple.

@Markiyan Hirnyk 

Workarounds? But these are bugs too!

@tomleslie 

Yes, it is Pi^2/8 with "probability 1".

It reduces to
J:=2*Int(arcsinh(cosh(x))-x, x=0..infinity);
identify(evalf(J));  #Pi^2/8

is(value(J)=Pi^2/8); #false

It seems to be hard to compute J symbolically.

@John Fredsted 

Theoretically it could be formulated as sequence of optimization problems:
find the vectors vi = [xi1,xi2,...,xi8] (i=1,...,n) satisfying the constraints ||vi||=1, ||vi-vj|| >= 1.

Then increase n until a solution does not exist.

But practically the number of the constraints n*(n+1)/2 is probably too large for n>50 say, to hope for a solution.

@John Fredsted 

Why do you say that? A double-check is always better than none!

Edit. I mean e.g. to use Maple to find a maximal set of vectors V on the sphere ||x||=r in R^8 such that ||x-y|| >= r, for x,y in V, x<>y.

@John Fredsted 

A simple inspection is enough. E.g. if both x,y have integer entries then x-y can have 1,2,3 or 4 nonzero entries, and in the case of a single nonzero entry, this must be +/-2. So ||x-y|| >= sqrt(2) [attained if x-y has 2 nonzero entries +/-1].

P.S. It would be nice to use Maple to find such a configuration.

I don't know much about sphere packing, but I don't understand why do you need a Maple check here, when mathematically this is almost obvious: for E8, if x,y are distinct "root" vectors, ||x||=sqrt(2), ||x-y|| >= sqrt(2).
Probably in one of the Leech constructions the situation is similar.

Optimization:-Maximize uses floats (just like DS), so, the true max could be 2 + exp(-10^99999999999).

@acer 

It seems that Or considers (here) only two arguments.

Conclusion: it's hard to live in the Maple world!

@acer 

OK, thank you, so

simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y<0,y=0,y>0);

works, and my infinity prediction was "highly exagerated".
I like the important ingredient Or(x<0,x=0,x>0).

Edit. I though it was a typo, but
simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y<0,y=0,y>0);
simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y=0,y<0);
indeed do not simplify in a fresh session, but do simplify later (due to remember tables)!