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Lagrangian Econ Problem

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the equations for the utility maximizing values for x and y

    U(x,y) = x^2 + y^2


    2. Relevant equations

    Budget constraint: I = PxX +Pyy

    L(x,y,\lambda ) x^2 + y^2 + \lambda (I - PxX - PyY)

    3. The attempt at a solution

    I got the three partial derivatives and set equal to zero:

    dL/dx = 2x - \lambda Px = 0
    dL/dy = 2y - \lambda Py = 0
    dL/d\lambda = I-PxX-PyY = 0

    Then i set the first two equal to each other to try and find x in terms of y

    2x = \lambda Px
    2y \lambda Py

    This results in x = PxY/Py


    But here's the problem...

    When I plug that into the last equation, i get stuck

    I get:

    I - PxX - Py(PyX/Px) = 0

    I dont know how to proceed from here algebraically. Normally I'd be able to cancel on some of the simpler problems. But I can't cancel the Px out from the denominator





    Any help would be greatly appreciated!!!
     
    Last edited: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2

    vela

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    Remember that ##p_x## and ##p_y## are just constants, so you have an equation of the form ##I - a x - b x = 0## where ##a = p_x## and ##b = p_y^2/p_x##. The simplest thing to do would be to multiply through by ##p_x##.
     
  4. Oct 1, 2012 #3

    Ray Vickson

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    The Lagrangian conditions give you x = (λ/2) a and y = (λ/2) b, where I use a and b in place of px and py. Using these in your constraint gives you everything you need.

    That is essentially the way in which most Lagrange multiplier problems are solved, although on rare occasions it is easier to use another method.

    RGV
     
    Last edited: Oct 1, 2012
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