@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

@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.