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Math Help - Critical points of a two-variable function.

  1. #1
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    Critical points of a two-variable function.

    Hi!
    I Can't seem to find the critical points of the function f(x,y) = x*y*ln(x^2 + y^2)

    I did the df/dx = y*[ (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) ]
    and the df/dy = x*[ (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) ]

    I equalled them both to 0, but now I'm stuck because I don't know how to solve the system:

    y*[ (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) ] = 0
    x*[ (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) ] = 0

    I have the solutions. There are 4 critical points and none of them is x=0 or y=0.
    But I don't know how to get to the solution.

    Thanks in advance!
    Tony
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  2. #2
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    Re: Critical points of a two-variable function.

    if x and y don't equal zero, then surely you can divide through by y in equation 1, x in equation 2. Then you have two things equal to 0... if you equate them what do you get?

    This is just a starting step...
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  3. #3
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    Re: Critical points of a two-variable function.

    I tried that but it got me nowhere. The solutions are:
    [ 1/(sqrt(2)*e); 1/(sqrt(2)*e) ], [ -1/(sqrt(2)*e); -1/(sqrt(2)*e) ],
    [-sqrt(e)/2 ; sqrt(e)/2] and [ sqrt(e)/2 ; -sqrt(e)/2 ].

    I don't see how I could get this from (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) = 0
    and (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) = 0

    I tried adding them, substracting, moving the 'ln' to the other side and raising them as power of 'e', but I can't seem to solve it.
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  4. #4
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    Re: Critical points of a two-variable function.

    Are your first solution supposed to be:

    x=y= \pm\frac{1}{\sqrt{2e}}

    Hopefully it is....

    divide through by x and y in the corresponding equation, you have two things that equal 0 then, so they equal each other...
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  5. #5
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    Re: Critical points of a two-variable function.

    Yes they are the two first solutions.

    I tried equalling them too but as I said I can't seem to move on. Please could you show me how you get these results?
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  6. #6
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    Re: Critical points of a two-variable function.

    your last 2 critical points are wrong i'm afraid, I have spent the last 10 minutes puzzling over those solutions, but now I have graphed it, and I have attained the correct solution.

    The set of solutions is...
    \left(\frac{1}{\sqrt{2e}},\frac{1}{\sqrt{2e}} \right),\left(-\frac{1}{\sqrt{2e}},-\frac{1}{\sqrt{2e}} \right),\left(\frac{1}{\sqrt{2e}},-\frac{1}{\sqrt{2e}} \right),\left(-\frac{1}{\sqrt{2e}},\frac{1}{\sqrt{2e}} \right),

    I will post working shortly
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  7. #7
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    Re: Critical points of a two-variable function.

    So you are right with your first 2 equations. Dividing through by y in the first, x in the second gives:

    \log(x^2+y^2)+\frac{2x^2}{x^2+y^2}=0=\log(x^2+y^2)  + \frac{2y^2}{x^2+y^2}

    Can you see where to go from here? if a+b=a+c then what do we know about b and c?
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  8. #8
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    Re: Critical points of a two-variable function.

    Yes, I've gotten to the point that x^2/(x^2 + y^2) = y^2/(x^2+y^2) if that's what you mean, but I can't go further.
    I don't know how to get the solutions [ 1/(sqrt(2)*e); 1/(sqrt(2)*e) ], etc. from it.
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  9. #9
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    Re: Critical points of a two-variable function.

    Never mind. I solved it. Your results were correct.
    Thanks for the help!
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