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Math Help - Another local extreme problem

  1. #1
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    Another local extreme problem

    I am facing another problem that I can't compute the extremes forDoh)

    f(xy)=ex(x2-y2)

    so far I have:

    fx= 2xex + ex(x2-y2)
    fy= -2yex
    fxx=2ex+2xex + ex(x2-y2) + 2xex
    fyy= -2ex
    fxy= -2yex

    How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

    Thanks for any help.
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  2. #2
    Behold, the power of SARDINES!
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    Re: Another local extreme problem

    Quote Originally Posted by MajMinor View Post
    I am facing another problem that I can't compute the extremes forDoh)

    f(xy)=ex(x2-y2)

    so far I have:

    fx= 2xex + ex(x2-y2)
    fy= -2yex
    fxx=2ex+2xex + ex(x2-y2) + 2xex
    fyy= -2ex
    fxy= -2yex

    How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

    Thanks for any help.
    If you set the partial with respect to x equal to 0 you get

    2xe^{x}+e^{x}(x^2-y^2)=0 \\ e^{x}(2x+x^2-y^2)=0


    Since the exponential is never zero his gives

    2x+x^2-y^2=0 \iff y=\pm \sqrt{x^2+2x}

    If we plug this into the 2nd equation we get


    2(\pm \sqrt{x^2+2x})e^{x}=0


    Again the exponential can never be zero so we get


    (\pm \sqrt{x^2+2x})=0 \iff x^2+2x=0^2 \iff x(x+2)=0

    So we get the two solutions

    x=0 \quad x=-2

    Can you finish from here?
    Thanks from MajMinor
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  3. #3
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    Re: Another local extreme problem

    Yes. Thank you very much! It's the little things that always trip me up.
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