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Math Help - [SOLVED] F(x,y) maximum problem

  1. #1
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    [SOLVED] F(x,y) maximum problem

    When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between 0 and  2\pi.

    If I have infinitely many critical points is there another method to solving this?

    f(x,y)=sin(x)+sin(y)+sin(x+y)

    f_x(x,y)=cos(x) + cos(x+y)
    f_y(x,y)=cos(y) + cos(x+y)

    Now f_x=0=f_y
    So when i set them equal-->
    cos(x) + cos(x+y)=cos(y) + cos(x+y)
    cos(x)=cos(y)
    x=y

    Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...

    EDIT: Here is a link to the graph I think it'll work: http://www.wolframalpha.com/input/?i...sin%28x%2By%29
    if not paste this in: graph z=sin(x)+sin(y)+sin(x+y) to get an idea of the graph.

    EDIT:Going to bed now, will check again around 12hrs from now...for what it is worth
    Last edited by snaes; October 7th 2009 at 07:30 PM. Reason: added link
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  2. #2
    MHF Contributor Calculus26's Avatar
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    since cos(x) + cos(x+y) = 0

    and since x = y

    cos(x) + cos(2x) = 0

    2cos^2(x) +cos(x) -1 = 0

    (2cos(x) -1)(cos(x)+1) = 0

    cos(x) = 1/2 cos(x) = -1

    I'm assuming the domain is the square [0,2pi] x[0,2pi] ?

    In which case x = pi/3, 5pi/3, pi

    Again since x = y

    (pi/3,pi/3) , (5pi/3,5pi/3) (pi,pi) are the critical points
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  3. #3
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    Quote Originally Posted by snaes View Post
    When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between 0 and  2\pi.

    If I have infinitely many critical points is there another method to solving this?

    f(x,y)=sin(x)+sin(y)+sin(x+y)

    f_x(x,y)=cos(x) + cos(x+y)
    f_y(x,y)=cos(y) + cos(x+y)

    Now f_x=0=f_y
    So when i set them equal-->
    cos(x) + cos(x+y)=cos(y) + cos(x+y)
    cos(x)=cos(y)
    x=y

    Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...
    From this is follows that if f_x= f_y then cos(x)= cos(y). It does NOT follow that x= y. You might also have y= 2\pi- x. So you actually have twice as many solutions as you thought!
    Those are the two diagonals of the square [0,2\pi]\times[0,2\pi].

    And, it does not follow that the two derivatives are equal to 0 for all y= x, only that they are equal two each other. Caculus26 showed that, using y= x in one of the equations, f_x= 0 or f_y= 0 you get (\pi/3,\pi/3) , (5\pi/3,5\pi/3), and (\pi,\pi).

    Do the same with y= 2\pi- x. Since x+y= x+ (2\pi- x)= 2\pi you have cos(x)+ cos(x+y)= cos(x)+ cos(2\pi)= cos(x)+ 1= 0 and that is true only for [tex]x= \pi[/itex]. Then y= 2\pi- \pi= \pi so that just gives the point (\pi, \pi) that Caculus26 got before- it is the center point of the square and lies on both diagonals.
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  4. #4
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    Ahh, I see. Thank you!
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