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March 28, 2017

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If the derivative can be thought of as a marginal revenue function for x units (in hundreds of items) sold, and the revenue for a company is given by the function.
R(x) = 30x^3 - 120x^2 + 500 for 0 _< x _< 100,

a. Sketch the graphs of the functions R(x) and R'(x) .

b. Find the number of units sold at which the marginal revenue begins to increase

  • Calculus - ,

    marginal revenue begins to increase when f'' changes sign.

    R'(x) = 90x^2 - 240x
    R"(x) = 180x-240 = 60(3x-4)

    So, marginal revenue begins to increase when x = 4/3

  • Calculus - ,

    r(x) = 30x^3 - 120x^2 + 500

    r ‘ (x) = 90x^2 - 240x

    0 = 90x^2 - 240x

    0 = 3x^2 - 8x

    0 = x(3x – 8)

    X = 0 or x = 8/3

    Or approx.

    X = 3

    Answer: x = 3

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