@snayperx 

 A simple procedure  MyPolyhedron  solves your problems. Required parameter  Faces  is the list of all faces of the polyhedron given as lists of its vertices. Optional parameters: 'plot' (by default), vertices, edges.

MyPolyhedron:=proc(Faces, opt:='plot')
local n, S;
n:=nops(Faces);
S:=seq(plottools:-polygon(Faces[i], color=khaki), i=1..n);
if opt='plot' then return plots:-display(S, scaling=constrained) else
if opt=vertices then return {op(map(op, Faces))} else
if opt=edges then {seq(seq({Faces[i,j], Faces[i,`if`(j<nops(Faces[i]), j+1, 1)]}, j=1..nops(Faces[i])), i=1..n)} fi; fi; fi;
end proc:  

Example of use:

Faces:=[[[2,1,0],[-2,1,0],[-2,-1,0],[2,-1,0]], [[2,1,0],[1,0,2],[2,-1,0]], [[2,1,0],[-2,1,0],[-1,0,2],[1,0,2]], [[-2,1,0],[-1,0,2],[-2,-1,0]], [[2,-1,0],[-2,-1,0],[-1,0,2],[1,0,2]]]:
MyPolyhedron(Faces);
MyPolyhedron(Faces, vertices);
MyPolyhedron(Faces, edges);
 

In Maple 2017.0 i get a solution without any problems:

restart;
Sys:={p=(w^2+v*w)/(1-2*u-w), q=u+w, r=w}:
solve(Sys, {u,v,w});
simplify(eval(Sys, %));  # Checking
 

@vv  Thank you for this improvement.

@vv  Maple copes with this example, if a little more help it (in  f1  we manually split  f  into 2 summands):

restart;
f := (3*(x+exp(x))^(2/3)+(x^2+3*x)*exp(x)+4*x^2)/((x+exp(x))^(2/3)*x);
f1:=3/x+``(((x^2+3*x)*exp(x)+4*x^2)/((x+exp(x))^(2/3)*x));
IntegrationTools:-Expand(int(f1, x));
normal(diff(%, x))  # Checking

@magiblot  From the plot it is clear that the maximum point lies in the third quadrant (x<0, y<0). Then by the mouse I just turned the plot (top view) and took a close point, lying in a triangle (x=-5, y=-5) as an initial point.  For  the minimum point everything is similar.

@Rock  Execute  restart  command at first.

@kuwait1 

KK := unapply(diff(n*x^2, x), x, n); 
KK(2, 2);

@jacob123 

Download calc.mw

@mqb  See the edited version above.

@vv  OK, probably you are right. If we give the same initial conditions for a slightly larger t (i took t = 0.1), then a solution exists, in this case f''(0.1) is very large:

l := t -> 0.5*tanh(0.5*t):
deq := diff(f(t), t)*l(t)*(diff(f(t), t, t)*l(t)+9.8*sin(f(t)))+diff(l(t), t)*(diff(f(t), t)^2*l(t)-9.8*cos(f(t))+4*(l(t)-0.5)) = 0, diff(f(t), t, t)=g(t):
sol := dsolve({deq, f(0.1) = 0, D(f)(0.1) = 0.1}, {f(t),g(t)}, numeric);
plots:-odeplot(sol,[t,f(t)],t=0.1..1);
eval(g(t),sol(0.1));
 

@Carl Love   It is better to take the exponent at x of 1.5 instead of 2:

plot(`if`(x>=0 and x<=0.982, 0.015*x^1.5*sin(280*x), NULL), x=0..1.2, -0.02..0.02, color=blue, size=[800,400], gridlines);

I saw two absolutely identical questions for the last 1-2 days and removed the last one as spam. It seems that the second question was automatically deleted. Probably I accidentally pressed the wrong button. I ask you to excuse me for this imprudence.

@John Fredsted 

getMinMax(53^3);

                                                3, 53

Should be   53, 53

See the thread with the same question  http://www.mapleprimes.com/questions/202504-Strange-Behavior-Of--Plotscontourplot

@gtbastos   Try:

restart;
M:=combinat:-permute([0$16,1$16],16):
S:=[seq(`if`(LinearAlgebra:-Modular:-Rank(3, Matrix(4,m, datatype= float[8]))>=3, m, NULL), m=M)]:  # The list of all such matrices, represented as the lists formed of the rows
nops(S);   # The total number of such matrices