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Lagrange multipliers

  1. Jun 27, 2003 #1

    Dx

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    Find max and min value…f(x,y,z) =3x+2y+z; x2 + y2+z2 = 1

    If g(x,y,z) = x2 + y2+z2 = 1 then what do I do next?

    I need help to further solve for this plz? I am horrible at math and dont understand lagrange multipiers so can anyone better explain it to me and help me solve for difficult problem.

    Dx
     
  2. jcsd
  3. Jun 27, 2003 #2

    Hurkyl

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    Well, my text has the following box for lagrange multipliers (interesting theorems / procedures are placed in boxes in the text):

    I imagine your text has something similar. Do you understand how to start step (a)?
     
  4. Jun 27, 2003 #3

    Dx

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    No, I dont understand this what-so-ever. I see one example in my book that wants to find the points of a rectangular hyperbola and its using partial Dx but I dont know how to perform step a could you plz help me.

    Dx
     
  5. Jun 27, 2003 #4

    Hurkyl

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    ∇f(x, y, z) means the gradient of f with respect to its three variables. The gradient is defined as the row vector:

    &nabla;f(x, y, z) = < &part;f(x, y, z)/&part;x, &part;f(x, y, z)/&part;y, &part;f(x, y, z)/&part;z >
    I.E. the first coordinate is the partial derivative with respect to (WRT) the first variable, the second coordinate is the partial derivative WRT the second variable, and so on for as many variabes as the function has.


    The equation in step (a) is, then, a vector equation, which we solve by setting corresponding coordinates equal:

    &nabla;f(x, y, z) = &lambda; &nabla;g(x, y, z)

    is

    < &part;f(x, y, z)/&part;x, &part;f(x, y, z)/&part;y, &part;f(x, y, z)/&part;z > = &lambda; < &part;g(x, y, z)/&part;x, &part;g(x, y, z)/&part;y, &part;g(x, y, z)/&part;z >

    is

    &part;f(x, y, z)/&part;x = &lambda; &part;g(x, y, z)/&part;x
    &part;f(x, y, z)/&part;y = &lambda; &part;g(x, y, z)/&part;y
    &part;f(x, y, z)/&part;z = &lambda; &part;g(x, y, z)/&part;z



    (BTW, if you also want me to explain the "why" behind this method, ask and I'll do so... but the "why" may be quite difficult to understand)
     
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