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Your company can make x-hundred grade A tires and y-hundred grade B tires a day, where 0≤x≤4 and y= (40-10x)/(5-x). The profit on each grade A tire is twice the profit on each grade B tire. Assuming all tires sell, what are the most profitable numbers of tire to make?

  • math -

    Profit = 2(x) + 1(40-10x)/(5-x)
    = 2x + (40-10x)/(5-x)
    = (2x^2 - 40)/(x-5)
    d(profit)/dx = [(x-5)(4x) - (2x^2 - 40)(1) ]/(x-5)^2
    = 0 for a max of profit

    4x^2 - 20x - 2x^2 + 40 = 0

    x^2 - 10x + 20 = 0
    x = 5 ± √5
    = appr 7.24 or appr 2.76
    but x must be between 0 and 4, so x = 2.76
    where x is in hundreds

    so for a max profit they should make 276 A tires
    and 552 B tires

    check my arithmetic,

  • math -

    (3,0),(-2,6),(1,5),(3,6)

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