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Math Help - Lagrange Argh...

  1. #1
    Member billym's Avatar
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    Lagrange Argh...

    Looking to find and classify critical points:

    f(x,y)=x^2+y^2+20

    subject to constraint:

    3x^2+4xy+6y^2=140

    So, let g(x,y)=3x^2+4xy+6y^2-140=0

    f_x+\lambda g_x=0 \Rightarrow 2x+\lambda (6x+4y)

    which gives: \lambda(6x+4y)=-2x\Rightarrow \lambda=-\frac{x}{3x+2y}

    f_y+\lambda g_y=0 \Rightarrow 2y+\lambda (4x+12y)

    which gives: \lambda(4x+12y)=-2y\Rightarrow \lambda=-\frac{y}{2x+6y}

    Thus:

    \frac{x}{3x+2y}=\frac{y}{2x+6y}

    And I'm stuck...
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  2. #2
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    Quote Originally Posted by billym View Post
    Looking to find and classify critical points:

    f(x,y)=x^2+y^2+20

    subject to constraint:

    3x^2+4xy+6y^2=140

    So, let g(x,y)=3x^2+4xy+6y^2-140=0

    f_x+\lambda g_x=0 \Rightarrow 2x+\lambda (6x+4y)

    which gives: \lambda(6x+4y)=-2x\Rightarrow \lambda=-\frac{x}{3x+2y}

    f_y+\lambda g_y=0 \Rightarrow 2y+\lambda (4x+12y)

    which gives: \lambda(4x+12y)=-2y\Rightarrow \lambda=-\frac{y}{2x+6y}

    Thus:

    \frac{x}{3x+2y}=\frac{y}{2x+6y}

    And I'm stuck...

    ...then you continue! Cross multiply and get: 2x^2 + 3yx - 2y^2 = 0 . This is a quadratic in x, and solving it we get 2(x + 2y)(x - y/2) = 0 ==> input these solutions for x now in g(x,y) and find y, and then x. I thinbk there are 4 different points, two of which are (2,4) and (-2, -4)...check this, though! I could easily be wrong...

    Tonio
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  3. #3
    Member billym's Avatar
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    ugh... I think my brain might is broken:

    for y=2x:

    3x^2+4x(2x)+6(2x)^2=140 \Rightarrow 35x^2=140 \Rightarrow x= \pm 2

    for y=-x/2 :

    3x^2+4x(-x/2)+6(-x/2)^2=140 \Rightarrow x = \sqrt {56}

    ?!!
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    Quote Originally Posted by billym View Post
    ugh... I think my brain might is broken:

    for y=2x:

    3x^2+4x(2x)+6(2x)^2=140 \Rightarrow 35x^2=140 \Rightarrow x= \pm 2

    for y=-x/2 :

    3x^2+4x(-x/2)+6(-x/2)^2=140 \Rightarrow x = \sqrt {56}

    ?!!

    Yes...any problem? Of course, should be +/- Sqrt(56), and then you must calcualte the corresponding values for y, but it's fine imo.

    Tonio
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  5. #5
    Member billym's Avatar
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    (2 \sqrt{14}, - \sqrt{14}), (-2 \sqrt{14},  \sqrt{14}) ?
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    Quote Originally Posted by billym View Post
    (2 \sqrt{14}, - \sqrt{14}), (-2 \sqrt{14}, \sqrt{14}) ?

    Yup...and now find the other pair, when y = 2x.

    Tonio
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