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Stationary points for surface

  1. May 15, 2009 #1
    Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
    [tex]\f(x,y)= xye^{-1/2(x^2+y^2)}[/tex]

    Is there any stationary points for this surface?
    Let z = f(x,y)
    I found the
    [tex]\delta{z}/\delta{x}[/tex] = [tex]y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}][/tex]

    and

    [tex]\delta{z}/\delta{x}[/tex] = [tex]x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}][/tex]

    I'm supposed to equate the two equations to get the stationary points (x , y, z) But I couldn't since I could only get [tex]y - x = y^2 - x^2[/tex] from the two equations and that's it.

    How do I proceed? Did i do anything wrong?
     
  2. jcsd
  3. May 15, 2009 #2

    tiny-tim

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    Hi custer! :smile:

    (type \partial not \delta for ∂ :wink:)
    You got y = x2 and x = y2, and you combined them by adding (or subtracting?) them …

    try substituting one into the other instead. :wink:
     
  4. May 15, 2009 #3
    I think it should be
    [tex]y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]= y(1- x^2) e^{-1/2(x^2+ y^2)}= 0[/tex]
    [tex]x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]= x(1- y^2)e^{-1/2(x^2+ y^2)}= 0[/tex]
     
  5. May 15, 2009 #4

    tiny-tim

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    oops!

    oops! you're right, I misread it :redface:

    ok … so you have y(1- x2) = 0 and x(1- y2) = 0 …

    solve the first equation (there'll be three solutions), and then substitute (each solution into the second equation). :smile:
     
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