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Partial Derivatives

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find point closest to origin xy2z3 = 2



    2. Relevant equations



    3. The attempt at a solution

    note, k = lagrange multiplier

    grad f = 2xi + 2yj + 2zk, k grad f = k(y2z3i + 2xyz3j + 3z2xy2k)

    k = 2xy-2z-3 = x-1z-3 = (2/3)z-1x-1y-2

    y = [tex]\sqrt{2x^2}[/tex]

    x = [tex]\sqrt{(y^2)/2}[/tex]

    z = [tex]\sqrt{(3y^2)/2}[/tex]

    Plug x and z into the original

    ([tex]\sqrt{(y^2)/2}[/tex])(y2)([tex]\sqrt{(3y^2)/2}[/tex])3 = 2

    I tried to simplify that, first i squared both sides

    (y2/2)(y4)((27y6)/8) = 4

    (27/16)y12 = 4

    y = (4(16/27))1/12

    y = (64/27)1/12 = 1.074569932

    anyone agree, diagree
     
  2. jcsd
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