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Thread: Finding and classifying all relative extrema

  1. #1
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    Finding and classifying all relative extrema

    The function I have to find the relative extreme for is f(x,y)= x^3 + y^3 + 3x^2 - 3y^2 the question didn't say it was subject to any constraints and the only ones we,be doing have been subject to the constraint x^2+y^2=1 so I'm not sure what to do for this one, thanks in advance
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    Re: Finding and classifying all relative extrema

    Try setting $\displaystyle \bigtriangledown f = 0 $

    That is, $\displaystyle \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

    And solve for $\displaystyle x$ and $\displaystyle y$.

    This should give you four points which can be classified using the Hessian matrix.
    Last edited by TwoPlusTwo; Oct 30th 2013 at 10:57 AM.
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    Re: Finding and classifying all relative extrema

    Quote Originally Posted by TwoPlusTwo View Post
    Try setting $\displaystyle \bigtriangledown f = 0 $

    That is, $\displaystyle \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

    And solve for $\displaystyle x$ and $\displaystyle y$.

    This should give you four points which can be classified using the Hessian matrix.
    Thank you, so this would lead to $\displaystyle \frac{\partial f}{\partial x} = $ 3x^2+6x and $\displaystyle \frac{\partial f}{\partial y} = $ 3y^2-6y correct? And since they both equal zero does this bean that the realitive extrema would just be x=0 x=2 y=0 and y=-2? And I'm still unsure what my prfessor meant when he said to classify the extrema
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    Re: Finding and classifying all relative extrema

    Quote Originally Posted by grandmarquis84 View Post
    Thank you, so this would lead to $\displaystyle \frac{\partial f}{\partial x} = $ 3x^2+6x and $\displaystyle \frac{\partial f}{\partial y} = $ 3y^2-6y correct? And since they both equal zero does this bean that the realitive extrema would just be x=0 x=2 y=0 and y=-2? And I'm still unsure what my prfessor meant when he said to classify the extrema
    The partial derivatives are correct. The four critical points are:

    $\displaystyle (0,0), (0,2), (-2,0), (-2,2)$

    Classifying a critical point means to determine whether it's a minimum, maximum or a saddle point.

    For this you can use the Hessian matrix

    $\displaystyle H = \left(\begin{array}{cc}f_{xx} &f_{yx} \\f_{xy} &f_{yy}\end{array}\right)$.

    Take it's determinant:

    $\displaystyle |H| = f_{xx}f_{yy} - f_{xy}f_{yx}$

    Then any given critical point is

    1. a local minimum if $\displaystyle |H|>0$ and $\displaystyle f_{xx}>0$ in the point,
    2. a local maximum if $\displaystyle |H|>0$ and $\displaystyle f_{xx}<0$ in the point,
    3. a saddle point if $\displaystyle |H|<0$ in the point.

    If $\displaystyle |H|=0$ the test is inconclusive.

    Using this I got:

    $\displaystyle (0,0)$ and $\displaystyle (-2,2)$ are saddle points.

    $\displaystyle (0,2)$ is a local minimum, with $\displaystyle f(0,2)=-4$

    $\displaystyle (-2,0)$ is a local maximum, with $\displaystyle f(-2,0)=4$

    I suggest you try it yourself and see if you get the same results.
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