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 ¨C 120x^2 + 500 f or 0 ¡Ü x ¡Ü 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.
To sketch the graphs of the functions R(x) and R'(x), we need to find the function for the derivative R'(x) first.
Given that the revenue function is R(x) = 30x^3 - 120x^2 + 500, we can find the derivative as follows:
R'(x) = d/dx (30x^3 - 120x^2 + 500)
= 90x^2 - 240x
Now, let's sketch the graphs:
a. Graph of R(x):
To sketch the graph of R(x), we plot points for different values of x within the given range (0 ≤ x ≤ 100). We can calculate the corresponding values for R(x) as follows:
When x = 0:
R(0) = 30(0)^3 - 120(0)^2 + 500 = 500
When x = 10:
R(10) = 30(10)^3 - 120(10)^2 + 500 = 30,500 - 12000 + 500 = 19,000
When x = 50:
R(50) = 30(50)^3 - 120(50)^2 + 500 = 3,750,000 - 300,000 + 500 = 3,450,500
When x = 100:
R(100) = 30(100)^3 - 120(100)^2 + 500 = 300,000,000 - 1,200,000 + 500 = 298,800,500
Plotting these points on a graph, we can draw a curve that represents the function R(x).
b. Graph of R'(x):
To sketch the graph of R'(x), we plot points for different values of x within the given range (0 ≤ x ≤ 100). We can calculate the corresponding values for R'(x) as follows:
When x = 0:
R'(0) = 90(0)^2 - 240(0) = 0
When x = 10:
R'(10) = 90(10)^2 - 240(10) = 9,000 - 2,400 = 6,600
When x = 50:
R'(50) = 90(50)^2 - 240(50) = 202,500 - 12,000 = 190,500
When x = 100:
R'(100) = 90(100)^2 - 240(100) = 900,000 - 24,000 = 876,000
Plotting these points on a graph, we can draw a curve that represents the function R'(x).
b. Finding the number of units sold at which the marginal revenue begins to increase:
To find the number of units sold at which the marginal revenue begins to increase, we need to find the x-coordinate where R'(x) changes from negative (decreasing) to positive (increasing).
Looking at the graph of R'(x), we can see that the x-coordinate where R'(x) changes sign is between x = 50 and x = 100. We can use a numerical method such as the bisection method or Newton's method to find the exact value.