Posted by **Wiz** on Thursday, April 4, 2013 at 12:30pm.

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 -
**Steve**, Thursday, April 4, 2013 at 1:29pm
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 -
**Wiz**, Sunday, April 7, 2013 at 4:54pm
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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