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Homework Help: Absolute Extrema

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data

    F(x,y)= sin(x)sin(y)sin(x+y) over the square 0< x< pi and 0< y< pi


    (The values for x and y should be from 0 to pi INCLUSIVE)

    2. Relevant equations



    3. The attempt at a solution

    partial derivative in terms of x = siny[cosxsin(x+y)+sinxcos(x+y)]
    you get y=0, pi because siny =0, but I don't know how to solve for the other solutions


    partial derivative in terms of y = sinx[cosysin(x+y)+ sinycos(x+y)]
    and you get x=0, pi because sinx=0

    and then I have the same problem again
     
  2. jcsd
  3. Mar 29, 2010 #2

    vela

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    Use a trig identity to simplify the contents in the square brackets.
     
  4. Mar 29, 2010 #3
    I used the trig identities sin(x+y)= sinxcosy+cosxsiny and cos (x+y)= cosxcosy - sinxsiny

    and with that, I get the equation the previous equation multiplied out which gives me
    =cosxsinxcosy +cos^2(x)siny +sinxcosxcosy-sin^2(x)siny
    =2cosxsinxcosy +(cos^2(x)+sin^2(x))(siny-siny)
    =2cosxsinxcosy +(1)(0)= 2cosxsinxcosy
    and I did the same process for the y derivative and got it to equal
    = 2cosxcosysiny

    that x=0,pi/2, pi and y = 0, pi/2, pi

    then when I go to find the values of F(x,y) at the nine possible extrema points, the value of F(x,y) is equal to 0.

    Because they all came out to 0, I doubt that they are actually the extrema and probably did something wrong in my solving.
     
  5. Mar 29, 2010 #4
    anyone?
     
  6. Mar 29, 2010 #5

    vela

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    I meant you should go the other direction, e.g.,

    cos x sin(x+y) + sin x cos(x+y) = sin(2x+y)
     
  7. Mar 29, 2010 #6
    what formula did you use to find sin(2x+y)?
     
  8. Mar 29, 2010 #7

    vela

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    cos a sin b + sin a cos b = sin(a+b) with a=x and b=x+y.
     
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