1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper


    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook