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Finding local extremes from stationary points

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the stationary points and local extreme values of f(x,y) = xy^2e^-([tex]\frac{x^2 + y^2)}{2}[/tex]



    2. Relevant equations
    You need to find the gradient for the function and then set it equal to 0. So then df/dy equals df/dx and you can solve for a solution set.


    3. The attempt at a solution
    df/dx = y^2(-x^2 + 1)*e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

    df/dy = y*x(-y^2 + 2)*e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

    So since e^-([tex]\frac{x^2 + y^2)}{2}[/tex] cannot equal zero, y^2(-x^2 + 1) = 0
    and y*x(-y^2 + 2) = 0

    So then, y^2(-x^2 + 1) goes to y^2 - x^2y^2 = 0 and then [tex]\pm[/tex]1 = x

    Plugging this into the dy/dx equation gives me the following x, y critical points
    (0,0) (1,-[tex]\sqrt{2}[/tex]) (1, -[tex]\sqrt{2}[/tex]) (-1, [tex]\sqrt{2}[/tex]) (-1, [tex]\sqrt{-2}[/tex])

    So then to find out if they are max or mins or saddle points, I solve for the second partials of fx, fy, fxy

    fxx=A = x(-1 + 2y^2 - x^2y^2) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

    fyy=C = y(x - 2y + 3yx) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

    fxy=B = y(2- 2x^2 + y^2 = Y2x^2) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]


    Now I use the equation D = AC - B^2

    Can someone check my math out? It seems that something is wrong somewhere. I know this is a huge problem so thanks for any help.
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2

    EnumaElish

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    The critical points are:
    (x, 0) for any x
    (-1, -sqrt 2)
    (-1, sqrt 2)
    (1, -sqrt 2)
    (1, sqrt 2)

    In each 2nd order derivative, the polynomial term should simplify to:
    x(-3 + x^2)y^2 ... for "xx"
    x(2 - 5 y^2 + y^4) ... for "yy"
    y(-1 + x^2)(-2 + y^2) ... for "xy"
     
    Last edited: Feb 27, 2008
  4. Feb 27, 2008 #3
    So I was mostly right other than that being (x,0). Now I just plug those points in for x and y in that D = AC - B^2 right? How would I do that with the (x,0) point?
     
  5. Feb 27, 2008 #4

    EnumaElish

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    Plug in y = 0, leave x as x.
     
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