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Determining Local Extrema with Lagrange

  1. Nov 5, 2008 #1
    [SOLVED] Determining Local Extrema with Lagrange

    1. The problem statement, all variables and given/known data

    Find local extram of f(x,y,z) = 8x+4y-z with constraint g(x,y,z) = x^2 + y^2 + z^2 = 9

    2. Relevant equations

    [tex]\nabla f(x,y,z) = \lambda g(x,y,z)[/tex]

    3. The attempt at a solution

    So I did the partial derivatives for F and G:

    [tex]\nabla f(x,y,z) = (8,4,-1)[/tex] [tex]\nabla g(x,y,z) = (2x,2y,2x)[/tex]

    Now I use Lagrange to get

    [tex]8 =2 \lambda x , 4 = 2\lambda y , -1 =2 \lambda z[/tex]

    Now I isolated [tex]\lambda = \frac{4}{x} = \frac{2}{y} =- \frac{1}{2z}[/tex]

    Now I re-write both y and z in terms of x and I get:

    [tex]y = \frac{x}{2}, z =- \frac{x}{8}[/tex]

    Then I put them back in to g(x,y,z) = g(x,x/2,x/8)

    and get:

    [tex]g(x,x/2,x/8) = x^2 + \frac{x^2}{4} +\frac{x^2}{64} = 9 \Rightarrow \frac{64x^2 + 16x^2 + x^2}{64} = 9 \Rightarrow \frac{81x^2}{64} = 9 \Rightarrow x = \sqrt{\frac{193}{27}}[/tex]

    but the book says x = 8/3

    Can anyone see what mistake did I make? Thank You.
     
    Last edited: Nov 5, 2008
  2. jcsd
  3. Nov 5, 2008 #2

    Office_Shredder

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    Yeah I do. It's tricky. Let's rewind to... the step before

    81x2/64 = 9

    So dividing both sides by 81 we get

    x2/64 = 1/9

    Multiplying both sides by 64 we get

    x2 = 64/9

    Try solving that one instead
     
  4. Nov 5, 2008 #3
    x = 8/3 which agrees with the book. Thank You.

    but why couldn't I multiply 9 by 64, and then divide it by 81?
     
  5. Nov 5, 2008 #4

    gabbagabbahey

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    You can, but 9*64/81=192/27=64/9 not 193/27
     
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