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These are answers submitted by dinonykus

_EnvExplicit:=false: r:=1/2; a:=Pi; maximize(sqrt(1-2*r*cos(t)+r^2) +sqrt(1-2*r*cos(t-a)+r^2),t=-Pi..Pi,location); did not produce the answer. On the other hand, restart: r:=1/2.; a:=Pi; ansMax:=Optimization:-Maximize(sqrt(1-2*r*cos(t*Pi)+r^2) +sqrt(1-2*r*cos(t*Pi-a)+r^2),t=-1..1); ansMin:=Optimization:-Minimize(sqrt(1-2*r*cos(t*Pi)+r^2) +sqrt(1-2*r*cos(t*Pi-a)+r^2),t=-1..1); did give a numerical answer. Are there other ways to solve such problems. Thanks a bunch.
I'll remember this one. Thank you so much.
That's a clever way to go about. So it seems Maple can't directly give exact answers? Thanks a lot anyway. I appreciate your help.
Yes, I have maple 10, which actually gave a better result:) > Optimization:-Maximize(t^2*(1-t)^2*(1+t^2)*(1+(1-t)^2), t = 0 .. 1); [0.0976562500000000000, [t = 0.500000000000000000]] But, is there a way to give the exact answer rather than a numerical one? I am thinking about a possiblity that the max value might be sqrt(5)/16, attained at t=sqrt(5)/2. It would be preferable to have an exact answer in such cases. Thanks a lot.
> maximize(t^2*(1-t)^2*(1+t^2)*(1+(1-t)^2), t = 0 .. 1); > maximize((1/2-s)^2*(1/2+s)^2*(1+(1/2-s)^2)*(1+(1/2+s)^2), s = (-1)/2 .. 1/2); > maximize((1/4-s^2)^2*((5/4+s^2)^2-s^2), s = (-1)/2 .. 1/2); > maximize((1/4-u)^2*((5/4+u)^2-u), u = 0 .. 1/4); The first three lines failed to give the maximum. The last line gave the maximum 25/256. The original function t^2*(1-t)^2*(1+t^2)*(1+(1-t)^2) simplifies to (1/4-u)^2*((5/4+u)^2-u) by setting u=(1/2-t)^2. My question is: Is there a simpler way to directly give the max without going through the pain of simplification/transformation? Thanks a lot.
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