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a City's transit authority serves 178500 communters daily when the fare is $1.90. Market research has determined that every penny decrease in the fare will result in 1,050 new riders. what fare will maximize revenue?

  • Algebra -

    revenue = #riders * fare
    = (178500+1050(190-x))(.01x) for x<=190

    that's just a parabola with vertex at x=180.

  • Algebra -

    number of penny decreases ---- n

    cost of fare = 190-n
    number of riders = 178500 + 1050n

    revenue = R = (190-n)(178500+1050n)
    P' = (190-n)(1050) + (178500 + 1050n)(-1)
    = 0 for a max of P

    178500 + 1050n = 1050(190-n)
    divide by 1050
    170 + n = 190-n
    2n = 20
    n = 10

    the fare should be 190+10 = 200
    or it should be $ 2.00

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