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
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) = 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
To sketch the graphs of the functions R(x) and R'(x), we need to follow these steps:
a. Plotting R(x):
1. Choose some values for x within the given range (0 ≤ x ≤ 100). For example, you can choose x = 0, 25, 50, 75, 100.
2. Substitute these x-values into the equation R(x) = 30x^3 - 120x^2 + 500 to find the corresponding y-values.
3. Plot the points (x, R(x)) on a graph.
4. Connect the points with a smooth curve.
b. Plotting R'(x):
1. Derive the function R(x) with respect to x to find R'(x). In this case, the derivative is found by differentiating each term:
R'(x) = 90x^2 - 240x.
2. Choose the same values for x as in step 1 from plotting R(x).
3. Substitute these x-values into the equation R'(x) = 90x^2 - 240x to find the corresponding y-values.
4. Plot the points (x, R'(x)) on the same graph as R(x).
5. Connect the points with a smooth curve.
Now, to find the number of units sold at which the marginal revenue begins to increase (where R'(x) is positive), we need to solve the equation R'(x) > 0:
90x^2 - 240x > 0
Factoring out x:
x(90x - 240) > 0
The inequality is satisfied when either:
1. x > 0 (since if x = 0, R'(x) = 0)
2. 90x - 240 > 0
Solving the second inequality:
90x - 240 > 0
90x > 240
x > 2.67
Therefore, the marginal revenue begins to increase when the number of units sold (x) is greater than 2.67.