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Math Help - lagrange

  1. #1
    Member i_zz_y_ill's Avatar
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    lagrange

    as your awake how do i solve this. local max and minima from origin to curve x^3+y^3-6xy=0 i can't do it. answer is minima x=0=y(nb dont know working for) and max x=y=3(again don't know workin for)
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  2. #2
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    What have you tried?

    You need to solve \bigtriangledown f(x,y)=0
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  3. #3
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    Quote Originally Posted by i_zz_y_ill View Post
    as your awake how do i solve this. local max and minima from origin to curve x^3+y^3-6xy=0 i can't do it. answer is minima x=0=y(nb dont know working for) and max x=y=3(again don't know workin for)
    first,.you need to find the critical point by
    Quote Originally Posted by pickslides View Post
    solve \bigtriangledown f(x,y)=0
    and then use the second derivative test to determine the maxima and minima. see : Maxima and Minima of Functions of Two Variables
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  4. #4
    MHF Contributor Calculus26's Avatar
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    See attachment

    Note to dedust and pickslides we are not looking for critical points--we want to optimize the distance fn subject to a constraint --a lagrange multiplier pblm
    Attached Thumbnails Attached Thumbnails lagrange-lagrange.jpg  
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  5. #5
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    Hello, i_zz_y_ill!

    I assume you know to set up Lagrnge multipliers.


    Find the local max and minima from origin to curve x^3+y^3-6xy\:=\:0

    Answer are: minimum (0, 0), maximum (3,3)

    We want to minimize the distance from point (x,y) to the origin.

    The distance function is: . d \:=\:\sqrt{x^2+y^2}

    For convenience, we can minimize the square of that distance: . D \:=\:x^2+y^2


    Our function is: . f(x,y,\lambda) \;=\;(x^2+y^2) + \lambda(x^3 + y^3 - 6xy)


    Set the partial derivatives equal to 0, and solve the system . . .

    . . \begin{array}{cccccc}f_x &=& 2x + 3\lambda x^2 - 6y &=& 0 &[1] \\<br /> <br />
f_y &=& 2y + 3\lambda y^2 - 6x &=& 0 & [2] \\<br /> <br />
f_{\lambda} &=& x^3 + y^3 - 6xy &=& 0 & [3] \end{array}


    \begin{array}{ccccc}\text{From [1], we have:} & \lambda &=& \dfrac{6y-2x}{3x^2} & [4] \\ \\[-3mm]<br />
\text{From [2], we have:} & \lambda &=& \dfrac{6x-2y}{3y^2} & [5] \end{array}


    Equate [4] and [5]: . \frac{6y-2x}{3x^2} \:=\:\frac{6x-2y}{3y^2} \quad\Rightarrow\quad 3x^3 - 3y^3 - x^2y + xy^2 \:=\:0

    Factor: . 3(x^3-y^3) - xy(x-y) \:=\:0 \quad\Rightarrow\quad 3(x-y)(x^2+xy+y^2) - xy(x-y) \:=\:0

    Factor: . (x-y)\bigg[3(x^2+xy+y^2) - xy\bigg] \:=\:0 \quad\Rightarrow\quad 3(x-y)(3x^2 + 2xy + 3y^2) \:=\:0

    From x-y \:=\:0, we have: . \boxed{y \:=\:x}\;\;[6]

    And 3x^2 + 2xy + 3y^2 \:=\:0 has no real factors.


    Substitute [6] into [3]: . x^3 + x^3 - 6x^2 \:=\:0 \quad\Rightarrow\quad 2x^3 - 6x^2 \:=\:0 \quad\Rightarrow\quad 2x^2(x-3) \:=\:0

    . . Hence, we have: . x \:=\:0,3 \quad\hdots\quad y \:=\:0,3


    Therefore, the critical points are: . (0,0),\;(3,3)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I could be wrong, but I think this is the Folium of Descartes.

    I has a leaf-shaped loop in Quadrant 1
    . . . \text{and approaches }x + y + 3\:=\:0 \text{ as an asymptote.}


    Code:
                      |
                      |       o   o
                      |   o
                      |o          o 
        \ o           |
          \           o          o
            \   o     |       o
        ------\-------o---o----------
                \     |
                  \   |o
                    \ |
                      \  
                      | \
                      |   \o
                      |     \
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  6. #6
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    Talking

    Quote Originally Posted by Calculus26 View Post
    See attachment

    Note to dedust and pickslides we are not looking for critical points--we want to optimize the distance fn subject to a constraint --a lagrange multiplier pblm
    ah,..sorry,.i didn't read the problem carefully,.just continue what pickslides did
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