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Total production function using Lagrange Multipliers

  1. Mar 1, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Attached as Question.jpg.


    2. Relevant equations
    Partial differentiation.
    Lagrange multiplier equation.


    3. The attempt at a solution
    Attached as MyWork.jpg.

    Is my work correct? I'm still not confident with myself for these problems and it would be great if someone could confirm if I did this problem correctly and also, the values (or value since I can ignore the negative because it makes no sense in the problem) I get seem rather strange and, knowing my teacher, he usually gives problems with numbers that work out well and his notes also have nicely fitting numbers.
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2012 #2

    Ray Vickson

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    You made a blunder: you started with
    [tex] L = 12 x^{3/4}\, y^{1/4} + \lambda (100 x + 180 y - 25200) [/tex] and wrote
    [tex] \partial L/\partial y = 12 x^{3/4}\, y^{-1/4} + 180 \lambda, [/tex]
    which is incorrect.

    RGV
     
  4. Mar 1, 2012 #3

    s3a

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    Actually, I don't think so. (If that sounds rude, I don't intend to.)

    Looking back at my work here, I see a coefficient of 3 and not 12 for L_y.

    Could you tell me if I am right or wrong about this please?
     
  5. Mar 1, 2012 #4

    Ray Vickson

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    You are wrong. Go back and look at what you wrote in the third line of your attachment.

    RGV
     
  6. Mar 1, 2012 #5

    s3a

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    I marked it in red to show what I am talking about in case we're mis-communicating. (It seems to be the third line too.)
     

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  7. Mar 1, 2012 #6

    Ray Vickson

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    I was not referring to the "3"; I was referring to the [itex] y^{-1/4}.[/itex]

    RGV
     
  8. Mar 1, 2012 #7

    HallsofIvy

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    Yes, you have "3" correctly. However, your power of y is wrong. 1/4- 1 is not equal to -1/4.
     
  9. Mar 1, 2012 #8

    s3a

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    Oh, yeah, oops. :shy:

    Is the current attached work correct?
     

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