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Thread: Maximum and minimum problem

  1. #1
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    Maximum and minimum problem

    Hi there. I've got this function $\displaystyle f(x,y)=(y-3x^2)(y-x^2)$, and I have to analyze what happens at $\displaystyle (0,0)$ in terms of maxims and minims. But what I actually have to proof is that theres a saddle roof at that point.

    Theres is a critical point at $\displaystyle (0,0)$. Lets see:

    $\displaystyle f(x,y)=(y-3x^2)(y-x^2)=y^2-4yx^2+3x^4$

    $\displaystyle f_x=-8yx+12x^3$
    $\displaystyle f_y=2y-4x^2$
    Its clear there that there is a critical point at $\displaystyle (0,0)$

    The determinant of the partial derivatives of second order at $\displaystyle (0,0)$
    $\displaystyle f_{xx}=-8y+36x^2$,
    $\displaystyle f_{xy}=-8x=f_{yx}$,
    $\displaystyle f_{yy}=2$

    $\displaystyle f_{xx}(0,0)=0$

    $\displaystyle \left| \begin{array}{ccc}0 \ 0 \\ 0 \ 2 \\ \end{array} \right| =0$

    Then I can't say anything from there. And actually, if I try at any line that passes through $\displaystyle (0,0)$ I would find a minimum. I know that it isn't a minimum, but I don't know how to prove it.

    Bye there.
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  2. #2
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    The second derivative test should do the trick.

    The second derivative test states:

    $\displaystyle D=D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$

    1. If D > 0 and $\displaystyle f_{xx}(a,b)$ > 0, then f(a,b) is a local min.
    2. If D > 0 and $\displaystyle f_{xx}(a,b)$ < 0, then f(a,b) is a local max.
    3. If D < 0, then f(a,b) is a saddle point.
    4. If D = 0, then the test is inconclusive.
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  3. #3
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    The thing is that the determinant gives zero.
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  4. #4
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    The "Taylor's series" for a function of two variables, f(x,y), at (0, 0), is $\displaystyle f(0, 0)+ f_x(0, 0)x + f_y(0, 0)y+ \frac{f_{xx}(0, 0)}{2}x^2+ \frac{f_{xy}(0,0)}{2}xy+ \frac{f_{yy}(0, 0)}{2}y^2$+ terms of higher degree. Since you have already determined that the first derivatives at (0, 0) are 0, you are saying that the Taylor's series, at (0, 0), is $\displaystyle f(0, 0)+ \frac{0}{2}x^2+ \frac{0}{2}xy+ \frac{2}{2}y^2$+ terms of higher degree= $\displaystyle f(0, 0)+ y^2$+ terms of higher degree.

    You can see from that that the "$\displaystyle y^2$" term has a positive coefficient and so the graph curves upward in the y direction. Since the other second degree terms have 0 coefficient you need to look at the third degree terms which means you need to look at $\displaystyle f_{xxx}(0, 0)$, $\displaystyle f_{xyy}(0, 0)$, etc. If any of them are negative, then you have a saddle point and there is no maximum or minimum at (0, 0). If all of them are positive, then there is a minimum at (0, 0, 0).
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