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Lagrange Multipliers

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Find max/min of f subject to constraint: x^2+y^2+z^1 = 1


    2. Relevant equations


    f(x,y,z) = 1/4*x^2 + 1/9*y^2 + z^2
    g(x,y,z) = x^2 + y^2 + z^2 - 1

    3. The attempt at a solution

    L = 1/4*x^2 + 1/9*y^2 + z^2 - λ(x^2 + y^2 + z^2 - 1)
    Lx = 2/4*x - λ*x*2
    Ly = 2/9*y - λ*2*y
    Lz = 2*z - λ*2*z
     
  2. jcsd
  3. Apr 10, 2013 #2

    ehild

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    Correct so far. What should be the partial derivatives at a minimum/maximum?

    ehild
     
  4. Apr 10, 2013 #3
    I don't know, the lecturer told us to solve the rest of the problem at home.
     
  5. Apr 10, 2013 #4

    ehild

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    Well, check your notes, what is the condition that a minimum or maximum exist?

    ehild
     
  6. Apr 10, 2013 #5
    All 3 derivates of L (Lx, Ly, Lz) have to equal zero, to find the λ, then the x/y/z for critical points of f(x,y,z)

    Lx = 2/4*x - λ*x*2 = 0
    Ly = 2/9*y - λ*2*y = 0
    Lz = 2*z - λ*2*z = 0
     
  7. Apr 10, 2013 #6

    ehild

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    OK, factorize the left hand sides.

    ehild
     
  8. Apr 10, 2013 #7
    2x(1/4 - λ) = 0
    2y(1/9 - λ) = 0
    2z(1 - λ) = 0

    EDIT: I'm in a bit of a funk, apologies! (lectures from 8am 'til 3pm)
     
  9. Apr 10, 2013 #8

    ehild

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    So you need some rest, without Maths... :smile:
    Anyway: you can omit the factors 2. All your equations are products, equal to zero. That means, one of the two factors must be zero in each equation. More than one factor containing lambda can not be zero, it would mean contradiction. Can all x, y,z equal to zero? Remember, you have the condition that x2+y2+z2=1.

    ehild
     
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