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Math Help - maximizing revenue

  1. #1
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    maximizing revenue

    A film processory has 10 pickup stations in Manhattan and each of them produces an average daily revenue of $500. A marketing study has shown that opening additional pickup stations will reduce the average daily revenue for each location by $50 for each new station. Financial considerations allow the company to consider opening no more than 10 additional stations. How many new stations should be opened to maximize revenue?

    (not sure how to set up the equations)
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  2. #2
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    Hello, jamesk486@

    I get a strange answer.
    Is there a typo in the problem?


    A film processory has 10 pickup stations in Manhattan
    and each of them produces an average daily revenue of $500.

    A study has shown that opening additional pickup stations will reduce
    the average daily revenue for each location by $50 for each new station.

    The company considers opening no more than 10 additional stations.

    How many new stations should be opened to maximize revenue?

    Let x = number of new stations to be opened.

    There will be (10+x) stations operating.

    Each will produce (500-50x) dollars in revenue.

    The total revenue is: . R \;=\;(10+x)(500-50x) \;=\;5000-50x^2


    To maximize revenue:

    . . R'\,=\,0 \quad\Rightarrow\quad -100x \:=\:0 \quad\Rightarrow\quad x \:=\:0


    Therefore, they should open no new stations.


    . . . They are already making maximum revenue!
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  3. #3
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    yup this is the full question
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