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Math Help - the Calculus is killing my grades

  1. #1
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    the Calculus is killing my grades

    calculus problem-please someone help me out not the best at this
    p=D(x)=125e^-0.001x ; p=S(x)=25e^0.001x

    The value of x at equilibrium is___________

    Find the consumer's surplus and the producers' surplus at the equilibrium price level for the given price demand and price supply equations. Round all values to nearest integer.
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  2. #2
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    Re: the Calculus is killing my grades

    In order to start this problem we must note that at equilibrium Demand=Supply and therefore D(X)=S(X)

    125 e^{-0.001x} = 25 e^{0.001x}
    e^{0.002x} = \frac{125}{25}
    e^{0.002x}=5

    0.002x = ln(5)
    x = \frac{ln(5)}{0.002}
    x = 1.6094

    D(1.6094) = 125e^{(-0.001(1.6094))} =124.80
    S(1.6094) = 25 e^{(0.001(1.6094))} = 25.04

    Equilibrium demand and price = q(e),p(e) = [ 124.80, 25.04]
    p(e)q(e) = (124.80)(25.04) = 3,124.99

    CS = 125 \int^1_0 e^{-0.001x} dx -p(e)q(e)

    \int e^{-0.001x} dx = -e^{-0.001x} /.001 = -1,000e^{-0.001x}
    F(x) = -125,000 e^{-0.001x}
    F(124.80) = -125,000e^{(-0.001(124.80))} = -110334.177
    F(0) = -125000
    F(124.80)-F(0) = -110334.177 - (-125,000) = 14,665.82
    14,665.82 - p(e)q(e) = 14,665.82- 3,124.99 = 11540.83
    CS = 11,540.83

    moving on to PS

    =p(e)q(e) - 25 \int e^{0.001x} dx
    \int e^{0.001x} dx = 1000e^{0.001x}
    25,000 e^{0.001x}
    Let F(x) = 25,000 e^{0.001x}
    F(25.04) = 25633.90
    F(0) = 25000
    F(25)-F(0) = 25633.90 - 25000 = 633.90
    p(e)q(e) = 3,124.99 -633.90 = 2,491.09

    PS = 2,491.09

    you have posted similar problems now try those and see if you can follow the general procedure i have demonstrated
    Last edited by TheIntegrator; May 3rd 2012 at 01:31 PM.
    Thanks from braveheel
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  3. #3
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    Re: the Calculus is killing my grades

    thanks for the help. Got 2 more for you math brains.
    A contaminated lake is treated with bactericide.The rate of increase in harmful bacteria t days after the treatment is given by the following function where N(t) is the number of bacteria per milliliter of water. Since dN/dt is negative, the count of harmful bacteria is decreasing.

    dN/dt=-2000t/(1+t^2) 0=too or less then t =too less then 8

    (A) Find the minimum value of dN/dt
    The minimum value of dN/dt is________bacteria/ml per day

    #2-Find the indefinite integal
    / x^5e^x^6 dx=
    use integers or factions for any numbers in the expression. use C as the arbitrary constant
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