You said: I know how to solve it using either FindMaximum or Langrage multiplier if I assign numerical values to T1, T2, R, Rthhot, Rthcold, and Z.

Please upload a worksheet with your numerical example.

@tomleslie Actually the book is not a bad one. It has a different purpose. 
From the  preface:

It is appropriate to stress here that this book is NOT for learning how to use Maple, but rather for learning how to write and construct computer programs. We expect that the general programming principles learned in this book will greatly help students in their other programming courses (e.g., in C, Pascal, FORTRAN) particularly in their ability to analyze and write complex algorithms.

@mmcdara The "directional expansion" works if f is in the class C^n (near [0,0,...,0]) and the result is a polynomial.
Otherwise, the expansion fails or is non-polynomial. mtaylor does basically the same but also checks the polynomiality.

For example,

f:=(x,y)->sin(x^2*y/(x^2+y^2)):
normal(eval(series(f(t*x,t*y),t=0),t=1)); # non-polynomial

because f is not differentiable at [0,0], (although continuous, provided that f(0,0)=0).

@nm Just differentiate (1) wirh respect to t and then take t=1. Using Maple:

f(t*x, t^p*y) = t^(p-1)*f(x,y):
eval(diff(%, t), t=1); p = solve(%, p);

See a nicer solution here.

@acer My answer contains a simplified explanation based on:
-- the information in the help ?simplify/size
-- the output generated by the simplify commands using printlevel:=40;
 For example, we see here:

--> enter \`simplify/size/size/object\`, args = _z1^2*_z2^2+_z1^2*_z3^2

--> enter \`simplify/size/size/object\`, args = _z1^2*(_z2^2+_z3^2)

@mmcdara In newer versions, the size option is by default in simplify.

@lcz Not like this. A compilable procedure is very restrictive. It cannot contain maple structures (except rtable with hardware entries) , so it must be written from scratch, starting e.g. from the adjacency matrix of the graph.

@Carl Love Maybe with a more flexible input:

CyclSum:= proc(f::{procedure,algebraic}, x::{indexed(posint),list(name)})
    local j,k,n,t,g;
    if x::list then t:=x; n:=nops(x)  else t:=op(0,x); n:=op(1,x) fi;
    if f::procedure then g:=f elif x::list then g:=unapply(f,x) else error "2nd param must be a list here" fi; 
    add(g(seq(t[1+irem(j+k-2, n)], j= 1..n)), k= 1..n)
end:

CyclSum((x,y)->x^y, [x,y,z]);

CyclSum(x^y, [x,y,z]);

CyclSum( () -> args[1]^p*args[2]^q, x[7]);

@Carl Love I wonder what would be a functional procedure for constructing a cyclic sum associated to an arbitrary expression, i.e.  Sum( f(x[s(1)], ..., x[s(n)]), s in Cycl(n) ) ,  where Cycl(n) contains the n cyclic permutations of {1, ..., n}.

[ Probably you will enjoy to write it :-) ]

For symmetric polynomials, any such polynomial can be expressed as a polynomial in the symmetric fundamental sums [probably you used the term "basis" for these].
For n=3 variables (let's restrict the question for this case), these are x+y+z, x*y+y*z+z*x and x*y*z.
They are cyclic (for n=3) but obvbiously only symmetric polynomials can be obtained this way.

Probably there is not a finite set of "fundamental" cyclic polynomials. But for an infinite set, it is not difficult to see that your set of polynomials  union {x*y*z}  forms a "fundamental" set. 

Do you mean a basis in the vector space of cyclic polynomials with rational coefficients (say)?
In this case the answer is yes, because your polynomials are linearly independent.

@C_R  Rouben forgot to include:

addcoords(z_cylindrical,[z, r,theta],[r*cos(theta),r*sin(theta),z]);

For a unique solution, you will need to define a well-order (see wiki). 

The attachment contains an image hint. (For the moment mapleprimes refuses to inline it).

pic1.zip