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Math Help - Stationary points for surface

  1. #1
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    Stationary points for surface

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

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

    and

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

    I'm supposed to equate the two equations to get the stationary points (X , Y) and substitute them into f(x,y) to get Z. But I could not since I could only get y - x = y^2 - x^2 from the two equations and that's it.

    How do I proceed? Did i do anything wrong?
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  2. #2
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    \left\{\begin{array}{ll}f'_x=y(1-x^2)e^{-\frac{1}{2}(x^2+y^2)}=0\\f'_y=x(1-y^2)e^{-\frac{1}{2}(x^2+y^2)}=0\end{array}\right.

    e^{-\frac{1}{2}(x^2+y^2)} > 0, so you're left with:

    \left\{\begin{array}{ll}f'_x=y(1-x^2)=0\\f'_y=x(1-y^2)=0\end{array}\right.
    Last edited by Spec; May 15th 2009 at 02:23 AM. Reason: Mixed up the partial derivatives
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  3. #3
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    Quote Originally Posted by Spec View Post
    \left\{\begin{array}{ll}f'_x=y(1-x^2)e^{-\frac{1}{2}(x^2+y^2)}=0\\f'_y=x(1-y^2)e^{-\frac{1}{2}(x^2+y^2)}=0\end{array}\right.

    e^{-\frac{1}{2}(x^2+y^2)} > 0, so you're left with:

    \left\{\begin{array}{ll}f'_x=y(1-x^2)=0\\f'_y=x(1-y^2)=0\end{array}\right.
    Sorry. I don't get what you mean. I thought e^{-\frac{1}{2}(x^2+y^2)} will only result between 0 and 1, including 0 and 1.
    e^{-\infty} = 0 and e^0 = 1

    btw, i think i'm confused
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  4. #4
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    Quote Originally Posted by custer View Post
    Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
    f(x,y)= xye^{-1/2(x^2+y^2)}

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

    and

    \delta{z}/\delta{x} = x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]
    Do you mean \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}?


    I'm supposed to equate the two equations to get the stationary points (X , Y) and substitute them into f(x,y) to get Z. But I could not since I could only get y - x = y^2 - x^2 from the two equations and that's it.
    No, you do not "equate the two equations", you set each expression equal to 0.

    How do I proceed? Did i do anything wrong?
    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
    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

    Since an exponential is never 0, those are easy to solve.

    I find 5 stationary points.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Do you mean \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}?


    No, you do not "equate the two equations", you set each expression equal to 0.



    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
    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

    Since an exponential is never 0, those are easy to solve.

    I find 5 stationary points.
    By equating them I mean they both equals to zero.
    Anyway, million thanks
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