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Math Help - Critical points - Functions of two variables

  1. #1
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    Critical points - Functions of two variables

    For function  f(x,y) = 2x^2y + xy^2 - 4xy + 3
    Find all stationary points and describe their nature.

    So this is what I did, first I took the two partial derivatives.

     f_x = 4xy + y^2 - 4y

     f_y = 2x^2 + 2xy - 4x

    Allow the partial derivatives to equal zero to equate the critical points.

    For  f_x I divided both sides by y (not sure if this was a mistake or not), and for  f_y I divided both sides by x.

    This left me with:

     4x + y - 4 = 0

    and

     2x + 2y - 4 = 0

    From here, do I allow the two equations to equal each other and equate from there, or is that just going to get me into a muddle?

    I can't seem to work out a simple method to finding values that satisfy both equations aside from guessing (embarrasing I know).

    Thanks in advance,
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  2. #2
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    Quote Originally Posted by U-God View Post
    For function  f(x,y) = 2x^2y + xy^2 - 4xy + 3
    Find all stationary points and describe their nature.

    So this is what I did, first I took the two partial derivatives.

     f_x = 4xy + y^2 - 4y

     f_y = 2x^2 + 2xy - 4x

    Allow the partial derivatives to equal zero to equate the critical points.

    For  f_x I divided both sides by y (not sure if this was a mistake or not), and for  f_y I divided both sides by x.

    This left me with:

     4x + y - 4 = 0

    and

     2x + 2y - 4 = 0

    From here, do I allow the two equations to equal each other and equate from there, or is that just going to get me into a muddle?

    I can't seem to work out a simple method to finding values that satisfy both equations aside from guessing (embarrasing I know).

    Thanks in advance,
    Don't divide both sides by y or x! What if they were zero. All you must do is solve one of your equations for either x or y and substitute it into the other one. Maybe this multivariate calc has got your mind making this harder than it is but this is just simple algebra.
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  3. #3
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    Quote Originally Posted by U-God View Post
    For function  f(x,y) = 2x^2y + xy^2 - 4xy + 3
    Find all stationary points and describe their nature.

    So this is what I did, first I took the two partial derivatives.

     f_x = 4xy + y^2 - 4y

     f_y = 2x^2 + 2xy - 4x

    Allow the partial derivatives to equal zero to equate the critical points.

    For  f_x I divided both sides by y (not sure if this was a mistake or not), and for  f_y I divided both sides by x.

    This left me with:

     4x + y - 4 = 0 .... (1)

    and

     2x + 2y - 4 = 0 .... (2)

    From here, do I allow the two equations to equal each other and equate from there, or is that just going to get me into a muddle?

    I can't seem to work out a simple method to finding values that satisfy both equations aside from guessing (embarrasing I know).

    Thanks in advance,
    By dividing you have you have missed the solution (0, 0).

    Solve equations (1) and (2) simultaneously to get one of the other solutions.
    Last edited by mr fantastic; November 17th 2008 at 10:10 PM. Reason: Changed some wording
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    By dividing you have you have missed the solution (0, 0).

    Solve equations (1) and (2) simultaneously to get the other solution.
    Thanks to both of you for helping. I've got two of the solutions so far, but I'm told that there are four solutions. From equations (1) and (2) I can see that I can't get another solution because they're both linear.

    This makes me think that the other two solutions [besides (0,0) and (2/3,4/3)] have one zero in them, ie: (0,y) and (x,0). In which case I would have to take the first piece of advice given and re-arrange my two formulae and sub them in.

    However my algebra skills are lacking today and I can't seem to seperate x or y in either of the formulae without dividing by y or x respectively.

    A pointer for the first step or two would be appreciated.

    PS, I think my mind's just in a bit of a fluster because I've been studying too many consecutive hours for my exam tomorrow. Sorry if I'm missing something really obvious.

    Thanks,
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  5. #5
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    Quote Originally Posted by U-God View Post
    Thanks to both of you for helping. I've got two of the solutions so far, but I'm told that there are four solutions. From equations (1) and (2) I can see that I can't get another solution because they're both linear.

    This makes me think that the other two solutions [besides (0,0) and (2/3,4/3)] have one zero in them, ie: (0,y) and (x,0). In which case I would have to take the first piece of advice given and re-arrange my two formulae and sub them in.

    However my algebra skills are lacking today and I can't seem to seperate x or y in either of the formulae without dividing by y or x respectively.

    A pointer for the first step or two would be appreciated.

    PS, I think my mind's just in a bit of a fluster because I've been studying too many consecutive hours for my exam tomorrow. Sorry if I'm missing something really obvious.

    Thanks,
    y = 0 comes from the equation leading to equation (1). Substitute into equation (2): x = 2.

    x = 0 comes from the equation leading to equation (2). Substitute x = 0 into equation (1): y = 4.

    So your missing solutions are (2, 0) and (0, 4).
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