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Absolute Max/Min

  1. Feb 28, 2010 #1
    Can anyone tell me the general procedure in doing the following procedure?

    [tex]f(x,y)=xy^2[/tex] with domain [tex]x^2+y^2\leq4[/tex]

    Find it's absolute max & absolute min.

    Okay, here is my thought procedure, tell me what I can fix.

    So I would basically say, find the partial derivatives with respect to x and y and set them equal to zero.

    [tex]f_x=y^2=0[/tex] [tex]f_y=2yx=0[/tex]

    so what's up? I plug that into the original equation? and then do the whole matrix thing to find if it's an absolute max or min? so point [tex](x,y)=(0,0)[/tex]

    Plug into the matrix [tex]\left(\begin{array}{cc}f_x_x&f_x_y\\f_x_y&f_y_y\end{array}\right)[/tex]

    But I don't know how I would go about considering the [tex]x^2+y^2\leq4[/tex], do I find the boundary point? What are those? [tex](x,y)=(2,0)=(0,2)=(-2,0)=(0,-2)[/tex] and then plug it into the original equation and then use

    Plug into the matrix [tex]\left(\begin{array}{cc}f_x_x&f_x_y\\f_x_y&f_y_y\end{array}\right)[/tex]

    Am I on the right track? Can someone show me some guidance?
  2. jcsd
  3. Feb 28, 2010 #2
    You use the Hessian matrix (what you did above) to classify critical points. But the extrema do not have to occur at critical points--they can also occur at the boundary. In this case, the boundary consists of a circle of radius 2. Think about how the function behaves on this circle...maybe rewrite in terms of angle and see what you find.
  4. Feb 28, 2010 #3


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    In addition to what TinyBoss told you, you don't really need the Hessian in this problem. You first locate the critical points in the interior (there are lots of them,no?). Once you figure out the extremes on the boundary, which don't necessarily have to be at the axis intercepts just because you like them, you just list the points in the original function xy2. Then you can eyeball them to see the absolute max and min.
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