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Math Help - extremes of function

  1. #1
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    extremes of function

    Hi.

    appoint local extremes of function f(x,y)=x^3 + y^3 -3xy... i got two points, one is (0,0) and the other is (1,1) - is that correct? I wonder, where is the minimum and where is the maxium?

    Thank you!
    Last edited by sasom; May 27th 2008 at 06:31 AM.
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  2. #2
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    am... i used partial derivate so i got:

    df/dx = 3x^2 - 3y
    df/dy = 3y^2 - 3x

    local ekstrems are:

    3x^2 - 3y = 0
    3y^2 - 3x = 0
    .
    .
    .
    x1 = 0, x2 = 1
    y1 = 0, y2 = 1

    and then, where is maximum and where is "saddle",... I did secodn derivate and I used Hessian matrix, so in point (0,0) this matrix is negative (-9, so this is saddle right? - is this minimum?) and in point (1,1) a got positive matrix (27 - is this maximum or what?). Is all correct or I did some mistakes?
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  3. #3
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    Hello, sasom!

    2. Appoint local extremes of function f(x,y)\:=\:x^3 + y^3 -3xy

    i got two points: (0,0) and (1,1) ... is that correct? . . . . Yes!
    You found the first partial derivatives, equated them to zero, and solved . . .

    . . \begin{array}{cccccccc}<br />
f_x &=& 3x^2 - 3y &=& 0 & \Rightarrow & y \:=\:x^2 & {\color{blue}[1]}\\<br />
f_y &=& 3y^2 - 3x &=& 0 & \Rightarrow & x \:=\:y^2 & {\color{blue}[2]}<br />
\end{array}

    Substitute [1] into [2]: .  x \:=\:(x^2)^2 \quad\Rightarrow\quad x^4 - x \:=\:0 \quad\Rightarrow\quad x(x^3-1) \:=\:0

    Hence: . x \:=\:0,1\quad\Rightarrow\quad y \:=\:0,1

    There are two critical points: . (0,0,0) \text{ amd }(1,1,-1)



    Where is the minimum and where is the maximum?
    You're expected to know the Second Partials Test . . .

    f_{xx} \:=\;6x\qquad f_{yy} \:=\:6y\qquad f_{xy} \:=\:-3

    Then: . D \;=\;\left(f_{xx}\right)\left(f_{yy}\right) - \left(f_{xy}\right)^2 \;=\;(6x)(6y) - (-3)^2\;=\;9(4xy-1)

    At (0,0)\!:\;\;D \:=\:9(0-1) \:=\:-9\quad\hdots\;\;\boxed{\text{saddle point at }(0,0,0)}

    At (1,1)\!:\;\;D \:=\:9(3) \:=\: +27\quad\hdots\text{ extreme point}
    . . Since f_{xx}(1,1) \:=\:+6\quad\hdots\boxed{ \text{minimum at }(1,1,-1)}

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  4. #4
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    Thank you!
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