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Math Help - Help with min/max when D = 0

  1. #1
    Junior Member
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    Help with min/max when D = 0

    f(x,y) = e ^ (-x^2-y^2)

    at point (0,0), D = 0. Therefore the normal min/max test fails.

    The problem asks to classify the critical point at the origin by visualizing the surface z = f (x,y)

    when:

    f(x,0) = e^(-x^2) which is 1 / e(x^2) which would be POSITIVE
    f(-x,o) = POSITIVE
    f(0,y) = POSITIVE
    f(o,-y) = POSITIVE

    Shouldn't this mean this is a local/global minimum since as we move away from the point the function increases?

    However, the book says (0,0) would be a global maximum. Can someone explain?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Feb 2009
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    Swampscott, MA
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    943
    Quote Originally Posted by messianic View Post
    f(x,y) = e ^ (-x^2-y^2)

    at point (0,0), D = 0. Therefore the normal min/max test fails.

    The problem asks to classify the critical point at the origin by visualizing the surface z = f (x,y)

    when:

    f(x,0) = e^(-x^2) which is 1 / e(x^2) which would be POSITIVE
    f(-x,o) = POSITIVE
    f(0,y) = POSITIVE
    f(o,-y) = POSITIVE

    Shouldn't this mean this is a local/global minimum since as we move away from the point the function increases?

    However, the book says (0,0) would be a global maximum. Can someone explain?
    f_x=-2xe^{-(x^2+y^2)}
    f_y=-2ye^{-(x^2+y^2)}
    f_{xx}=2e^{-(x^2+y^2)}(2x^2-1)
    f_{yy}=2e^{-(x^2+y^2)}(2y^2-1)
    f_{xy}=4xye^{-(x^2+y^2)}

    At (0,0), f_{xx}=f_{yy}=-1 and f_{xy}=0 so f_{xx}f_{yy}-f_{xy}^2=1. This means we have either a local min or a local max (not a saddle point). But since f_{xx}(0,0) is negative, we can conclude that it is a local max.
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