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.

Question

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

Answer 1

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

Answer 2

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