1. The biggest cake cut in half and give to the first and the second persons. From the second cake cut off  7/32  part (as a sector) and give together with the least cake to the third person . The remaining  25/32  of the second cake goes to fourth person.

2. The general case is a challenge, and I still don't know how to solve it.

Test:=f->`if`((depends(op(1,f),x) and not depends(op(1,f),{y,z})) or (depends(op(2,f),y) and not depends(op(2,f),{x,z})) or (depends(op(3,f),z) and not depends(op(3,f),{x,y})), false, true);

Example of use:

fff:=[theta[1]*y+theta[2]*z,theta[5]*y,theta[10]*x+theta[11]*y+theta[12]*z+theta[14]*x*y+theta[15]*x*z+theta[16]*y*z+theta[17]*x^2+theta[18]*y*y+theta[19]*z*z+theta[20]]:

Test(fff);

                                                                         false

Edited.

May be it should be  (x-y)^2+(y-z)^2+(z-x)^2 = 3 . This is the infinite right circular cylinder of radius 1 whose axis is the line with parametric equation  x=t, y=t, z=t  (it is easy to prove).  Therefore your surface can be obtained by rotating around the axis  x=-t, y=t, z=t  of the standard cylinder  x^2+y^2=1  by the angle  arccos(1/sqrt(3)) :

C:=plot3d([cos(t),sin(t),h], t=0..2*Pi, h=-2..2, axes=normal, scaling=constrained, lightmodel=light1):

plottools[rotate](C, arccos(1/sqrt(3)), [[0,0,0], [-1,1,0]]);

In your example, always  R2 = 0  because  p=0 . Therefore, for any  n  you always get the same plot.  For the example I have taken  p=1 .

The plotting for  n=1, 6, 11 :

restart;

with(plots):

setoptions(title = `Family Plot`, axes = boxed):

pr := 0.71: p := 1: q := 0: b := 0: l := 0: s := 0: m := 0:

for i from 1 to 3 do

n:=1+5*(i-1);

R1 := 2.*n/(1+n);

R2 := 2.*p/(1+n);                                    

sol[i]:= dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+.71*f(eta)*(diff(theta(eta), eta))-.71*(diff(f(eta), eta))*theta(eta)*R2 = 0, f(0) = 0, (D(f))(0) = 1.8+b*((D@@2)(f))(0), (D(f))(18) = 0, theta(0) = 1+s*(D(theta))(0), theta(18) = 0], numeric, method = bvp[midrich]);

od:

display([seq(odeplot(sol[i], [eta, theta(eta)], eta=0..5, color = [red,blue,gold][i], axes = boxed),i=1..3)]);

It can be done as in the Help.

Example:

df := DataFrame( < 1, 2, 3, 4; 5, 6, 7, 8; 9, 10, 11, 12 >, 'rows' = [ 'a', 'b', 'c' ], 'columns' = [ 'A', 'B', 'C', 'D' ] );

df['a'..'b', 'B'..'D'];  # Removed third row and first column

eq2 := -4*A[2]*cos(2*x)-16*A[4]*cos(4*x)-36*A[6]*cos(6*x)-64*A[8]*cos(8*x)+a*A[0]+cos(8*x)*a*A[8]+cos(6*x)*a*A[6]+cos(4*x)*a*A[4]+2*cos(2*x)*q*A[0]+cos(2*x)*a*A[2]+q*A[8]*cos(6*x)+q*A[8]*cos(10*x)+q*A[6]*cos(4*x)+q*A[6]*cos(8*x)+q*A[4]*cos(2*x)+q*A[4]*cos(6*x)+q*A[2]*cos(4*x)+q*A[2]:

select(t->not has(t, cos), eq2);

                                                                      a*A[0]+q*A[2]

Try

simplify(A1^2+B1^2-CC1^2, size);

If I understand correctly, you want to both animations built simultaneously, rather than sequentially. To do this, remove the option  insequence = true . Also the parameters  b  and  J  should be specified in advance.

Example:

b:=1: J:=2:

tau[1] := h-1+(4/3)*(1/sqrt(3)-h+1)*sqrt(3)*(sqrt(3)*((1/4)*J^2-z^2)/(J^2*b));

tau[2] := (4/3)*sqrt(3)*((1/4)*h^2-z^2)/(h^2*b);

plot2 := plots[animate](plot, [tau[1], z = -(1/2)*J .. (1/2)*J, color = red, legend = shear*stress], h = 1 .. 1+1/sqrt(3)):

plot3 := plots[animate](plot, [tau[2], z = -(1/2)*h .. (1/2)*h, color = blue, legend = shear*stress], h = 0 .. 1):

plots[display]([plot2, plot3]);

Should be

factor(x^2-2*a*x+a^2);

                                                     (a - x)^2

Addition: In standard interface in 2d math you can do a space between  a  and  x . Maple regards a space between two symbols as a multiplication sign.  ax  in your code is just a new symbol, not  a*x . A space between a number (numeric type) and a symbol you do not necessarily have to do. For example in 2d math,  2a  is the same as 2 a  or  2*a 

V_matrix:=<seq(V[i], i=1..100)>;

You can solve the equation numerically for specific values of parameters and an initial condition.

Example:

eq:=diff(y(x),x) - (Q - x*p0*(exp(alpha-beta*y(x)))/(1+exp(alpha-beta*y(x))))^2=0:
sol:=dsolve({eq, y(0)=0}, numeric, parameters=[Q,p0,alpha,beta] );
sol(parameters=[1,2,3,4]);
plots[odeplot](sol, [x,y(x)], x=0..5);

 Addition. `Series` solution (above) fits only near x=0. Compare: 

restart;
eq:=diff(y(x),x) - (Q - x*p0*(exp(alpha-beta*y(x)))/(1+exp(alpha-beta*y(x))))^2=0:
P:=[Q,p0,alpha,beta]:
sol1:=dsolve({eq,y(0)=0}, numeric, parameters=P):
sol1(parameters=[1,2,3,4]):
dsolve([eq, y(0) = 0], y(x), type = 'series'):
sol2:=convert(%, polynom):
sol2:=eval(rhs(sol2),P=~[1,2,3,4]);

plots[display](plots[odeplot](sol1,[x,y(x)], x=0..5, color=red, legend=`Numerical solution`), plot(sol2, x=0..5, color=blue, legend=`Solution by series`), view=[0..5,-1..1.2]);

In  #2  should be

convert~(L, string);

                               ["Norman.Mailer", "Richard.Brautigan"]

Even simplier example:

g:= a -> int(x+a, x=a..2*a):

g(1);   # OK

eval(g(x), x=1);  # The error ( a premature calculation)

eval(g(z), z=1);  # OK

eval('g(x)', x=1);  # OK (a workaround)

V:=Int(x*sqrt(2*x^4+3), x);

value(V);  # The result (a direct calculation)

IntegrationTools[Change](V, u = sqrt(2)*x^2);  # Calculation by a change

value(%);

combine(expand(eval(%, u = sqrt(2)*x^2)));   # The result

Int  is an inert form of an integral.

Here is a simple example of the equation with the explicit solution, in which the same error:

dsolve({diff(y(x),x)*(x-1)^2+1=0, y(0)=0});

sol:=dsolve({diff(y(x),x)*(x-1)^2+1=0, y(0)=0}, numeric);

sol(2);

 It is obvious that the curve going from point (0,0) can not be continuously extended further right of  x = 1