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March 24, 2017

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Suppose the function p= -1/8 x + 100 (0 grtr&= x ~12) relates the selling price p of an item to the quantity x that is sold. Assume that p is in dollars. What is the maximum revenue possible in this situation?

  • pre-calc - ,

    Suppose the function p= -1/8 x + 100
    (0 grtr&= x ~12) relates the selling
    price p of an item to the quantity x
    that is sold. Assume that p is in
    dollars. What is the maximum
    revenue possible in this situation?

  • pre-calc - ,

    You should use the formula for revenue R(x)=p*q.
    In this case q is x. So you do R(x)= -1/8 x + 100(x). This results in R(x)= -1/8x^2 + 100x.
    To find the max, set this to 0 and complete the square:
    0=-1/8(x^2-800x)
    0=-1/8(x-400)^2+20000
    20,000 is your answer

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