Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
If the derivative can be thought of as a marginal revenue function for x units (in hundreds of items) sold, and the revenue for...Asked by Wiz
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
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
Answers
Answered by
Steve
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
R'(x) = 90x^2 - 240x
R"(x) = 180x-240 = 60(3x-4)
So, marginal revenue begins to increase when x = 4/3
Answered by
Wiz
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.