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Math Help - Calculus 3 - Optimization without Constraints

  1. #1
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    Unhappy Calculus 3 - Optimization without Constraints

    The behavior of a function can be complicated near a critical point where D=0. Suppose that f(x,y)= x^3 - 3xy^2.
    a.) Show that there is one critical point at (0,0) and that D=0 at that point.
    b.) Show that the contour for f(x,y)=0 consists of three lines intersecting at the origin where f alternates from positive to negative. Sketch a contour diagram for f near 0.

    I already did part a, but I have no idea how to even start part b.
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  2. #2
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    Quote Originally Posted by alyssalynnx38 View Post
    The behavior of a function can be complicated near a critical point where D=0. Suppose that f(x,y)= x^3 - 3xy^2.
    a.) Show that there is one critical point at (0,0) and that D=0 at that point.
    firstly solve for \nabla f = 0
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  3. #3
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    Like I said, I already did part a.

    fx = 3x^2 - 3y^2
    fy = -6xy
    fxx = 6x - 3y^2
    fyy = -6x
    fxy = fyx= -6y

    If you set fx and fy equal to 0, both x and y are 0 because xy = 0 and x^2 + y^2 = 0

    D = fxx*fyy - (fxy)^2
    D = (6x - 3y^2)*(-6x) - (-6y)^2
    D(0,0) = (6*0 - 3*0^2)*(-6*0) - (-6*0)^2 = 0

    So, there is a critical point at (0,0) and D = 0. And that's as far as I was able to get.
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