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Math Help - Critical Points

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    Critical Points

    I am doing my test corrections and I have all of them except one.

    Find the critical points if the function: f(x,y) = 1/3x^3 + 1/2y^2 + xy+1

    So far I have one crit point of (0,0), and I have the both partial derivative but I am not sure what other points there are. The partials that I got are:
    f'(x) = x^2 + y
    f'(y) = y + x

    Am I right so far? If so could somebody help me out there rest of the way? I am not familuar with the math text thing but I am trying to learnn
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    Quote Originally Posted by jazz836062 View Post
    I am doing my test corrections and I have all of them except one.

    Find the critical points if the function: f(x,y) = 1/3x^3 + 1/2y^2 + xy+1

    So far I have one crit point of (0,0), and I have the both partial derivative but I am not sure what other points there are. The partials that I got are:
    f'(x) = x^2 + y
    f'(y) = y + x

    Am I right so far? If so could somebody help me out there rest of the way? I am not familuar with the math text thing but I am trying to learnn
    ok so far

    now set each of those to zero.

    so we have x^2 + y = 0 and x + y = 0

    from the second equation, y = -x. plug this into the first equation, we get,

    x^2 - x = 0 \implies \boxed{x = 0} \text{ or } \boxed{x = 1}

    when x = 0, y = 0

    when x = 1, y = -x = -1

    so our critical points are (0,0) and (1,-1)
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    Now the second part of the question is to find if those points are relative minima, relative maxima, or a saddle point? Using the second derivative test I get no conclusion for (0,0) and (1,-1) is a relative min. Is that correct so far?
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    Quote Originally Posted by jazz836062 View Post
    Now the second part of the question is to find if those points are relative minima, relative maxima, or a saddle point? Using the second derivative test I get no conclusion for (0,0) and (1,-1) is a relative min. Is that correct so far?
    yes
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    I am not sure how to get a conclusion from the point (0,0).
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    Quote Originally Posted by jazz836062 View Post
    I am not sure how to get a conclusion from the point (0,0).
    is it not enough to say the test gives no conclusion? your professor would expect you to find it? we have a degenerate critical point here. you were taught how to deal with those?

    if my memory serves me correctly, you can go through the task of factorizing f(x,y) (it is quadratic in y) and finding where it is positive and negative, and use that to determine how the function behaves. if this is for a calc 3 class, i doubt you would be required to go that far though
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    I scoured my text book and I think I found something. The discriminant = f(xx)*f(yy) - f(xy)^2
    That means for this:
    D = (2x)(1) - (1^2) = 2x-1 = 2(0) - 1 = -1
    Therefore (0,0) is a saddle point. Am I right?
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    Quote Originally Posted by jazz836062 View Post
    I scoured my text book and I think I found something. The discriminant = f(xx)*f(yy) - f(xy)^2
    That means for this:
    D = (2x)(1) - (1^2) = 2x-1 = 2(0) - 1 = -1
    Therefore (0,0) is a saddle point. Am I right?
    yes, that is correct

    how did we get D = 0 earlier?
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  9. #9
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    I don't even know.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jazz836062 View Post
    I don't even know.
    evidently, we're both tired. well, i know i am, that's my excuse. i don't know what yours is
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