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Math Help - Lagrange multiplier problem

  1. #1
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    Lagrange multiplier problem

    Hello I'm stuck on this problem.

    P(L,K) = 100L^0.25 * K^0.75
    P = production output
    L = labour
    K = capital
    Suppose that labour costs $40 per unit and capital costs are $60 per unit. Units lagrange multipliers, find the least expensive way of producing 10 000 units of output

    This is what i've done so far.
    C(L,K) = 40L + 60K P(L,K) = 100L^0.25 x K^0.75 = 10 000
    C = λP
    C = 40i + 60j P = 25λ(K/L)^0.75 i + 75λ(L/K)^0.25 j
    40 = 25λ(K/L)^0.75 60 = 75λ(L/K)^0.25
    2(L/K)^0.25 = (L/K)^0.75
    2L = K

    P(L,2L) = 100L^0.25 x 2L^0.75 = 10 000
    10 000 = 200L
    L = 50

    C(L,2L) = 40(50) + 60(100) = 8000

    The examples I did in class were much different, they had values +/- for x, and you just use the - for the min.
    I'm lost, I'm not sure if this is a min or a max.

    Anyhelp will be greatful. =)
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  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
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    Florida
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    Note you have a small mistake

    your term P(L,2L) = 100L^0.25 x 2L^0.75 = 10 000


    should be P(L,2L) = 100L^0.25 x (2L)^0.75 = 10 000


    2^(.75)L =100

    L = 100/2^(.75) = 59.5


    K = 119

    This has to be a minimum since there is no max

    Consider

    C(L,K) = 40L + 60K and P(L,K) = 100L^0.25 x K^0.75 = 10 000

    as L goes to 0 k goes to infinity and C goes to infinity

    similarly if k goes to 0

    Also to convince yourself solve for k in terms of L and plug this into C and graph--again you'll see this is a minimum
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