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Math Help - Absolute extrema of a 2-variable function

  1. #1
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    Absolute extrema of a 2-variable function

    So I have to find the absolute extrema of f(x,y)=xy on D={(x,y)|x^2+2y^2<=4}

    I took partial derivatives of xy to find a critical point (0,0). Now I have to find the critical points on the boundary. This is where x^2+2y^2=4. I tried following this (Extrema for Multivariable Functions - Classwiki) so I rearranged it into x=sqrt(4-2y^2) and subbed it into f(x,y).

    I now have f(x,y) = y*sqrt(4-2y^2). I'm not sure where to go from here. The example above is a bit unclear but I think I make that = 0. This means y = 0 or +-sqrt(2). But if you put those into f(x,y), they all come out as 0. So I have no idea if what I'm doing is right or if the method is even right. Any help would be appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by someguy12345 View Post
    So I have to find the absolute extrema of f(x,y)=xy on D={(x,y)|x^2+2y^2<=4}

    I took partial derivatives of xy to find a critical point (0,0). Now I have to find the critical points on the boundary. This is where x^2+2y^2=4. I tried following this (Extrema for Multivariable Functions - Classwiki) so I rearranged it into x=sqrt(4-2y^2) and subbed it into f(x,y).

    I now have f(x,y) = y*sqrt(4-2y^2). I'm not sure where to go from here. The example above is a bit unclear but I think I make that = 0. This means y = 0 or +-sqrt(2). But if you put those into f(x,y), they all come out as 0. So I have no idea if what I'm doing is right or if the method is even right. Any help would be appreciated.
    For the extrema on the boundary switch to polars.

    CB
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  3. #3
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    The lecturer put an example up and it turns out we're supposed to use Lagrange Multipliers. I got an absolute minimum in D at f=0 and an absolute maximum in δD (the boundary) at f=4/3.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by someguy12345 View Post
    The lecturer put an example up and it turns out we're supposed to use Lagrange Multipliers. I got an absolute minimum in D at f=0 and an absolute maximum in δD (the boundary) at f=4/3.
    Lagrange multipliers will work, but are unnecessary for a two variable problem with one equality constraint.

    CB
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