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Finding max and min of a function of several variables

  1. Oct 18, 2009 #1
    Finding max and min of a function

    1. The problem statement, all variables and given/known data

    Find min , max , and saddle points of the function :

    f(x,y) = e^x*cos(y)

    f_x = e^x*cos(y)
    f_xx = e^x*cos(y)
    f_xy = -e^x * sin(y)

    f_y = -e^x * sin(y)
    f_y = -e^x * cos(y)

    First finding the critical points :

    f_x = e^x*cos(y)
    f_y = -e^x * sin(y)

    I know that e^x will never be 0 , so either cos(y) or sin(y) has to be 0;

    cos(y) = 0 , when y = (2n+1)*pi/2
    sin(y) = 0 , when y = n*pi

    where n is a natural number.

    how would I proceed next ?
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Dick

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    At this point I would start considering the possibility that f(x,y) doesn't have any critical points.
     
  4. Oct 18, 2009 #3
    Yea you are right, in the back of the book it says NONE, but how do I justify that?
     
  5. Oct 18, 2009 #4

    Dick

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    You want cos(y)=0 AND sin(y)=0 to have a critical point. Look at the conditions you've figured out for that possibility. Can they ever BOTH be true?
     
  6. Oct 18, 2009 #5
    cos(y) = 0 , when y = (2n+1)*pi/2
    sin(y) = 0 , when y = n*pi


    (2n+1)*pi/2 = n*pi

    pi*n + pi/2 = n*pi
    pi*n - pi*n = -pi/2

    0 = -pi/2

    so it seems like there is no solution or no critical point.

    Thanks a lot Dick for your help. I really mean it.
     
  7. Oct 18, 2009 #6

    Dick

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    Very welcome. But you've got a technical flaw. You shouldn't assume both n's are the same. You want to look for solutions of (2m+1)*pi/2 = n*pi where n and m might be different integers. That's (m-n)pi+pi/2=0. Still no solution, of course. Just a heads up.
     
  8. Oct 18, 2009 #7
    Amazing, smart people bring smile to my face.
     
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