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 understand how they are related. R(x) represents the revenue function, while R'(x) represents the derivative of the revenue function, which can be thought of as the marginal revenue function.
To sketch the graph of R(x), we can start by identifying the key points and intervals:
1. The domain of R(x) is given as 0 ≤ x ≤ 100 because the company sells a maximum of 100 units.
2. The revenue function R(x) = 30x^3 - 120x^2 + 500 is a cubic polynomial.
To find the key points of R(x), we can find the x-values where the derivative is equal to zero. In this case, we need to find when R'(x) = 0 to identify any local extremes. To find the derivative R'(x), we can differentiate R(x) with respect to x:
R'(x) = d/dx(30x^3 - 120x^2 + 500)
= 90x^2 - 240x
Now, to find when R'(x) = 0, we set the derivative equal to zero and solve for x:
90x^2 - 240x = 0
Factoring out x, we get:
x(90x - 240) = 0
Setting each factor equal to zero, we have:
x = 0 or 90x - 240 = 0
Solving for x in the second equation, we find:
90x = 240
x = 240/90
x = 8/3
So the derivative R'(x) has one critical point at x = 0 and x = 8/3.
Now let's find the corresponding y-values for these x-values:
R(0) = 30(0)^3 - 120(0)^2 + 500 = 500
R(8/3) = 30(8/3)^3 - 120(8/3)^2 + 500 = 832
Now we have the key points for R(x): (0, 500) and (8/3, 832).
We can now plot these key points and any additional points to sketch the graph of R(x) within the given domain of 0 ≤ x ≤ 100.
To sketch the graph of R'(x), we can use the same key points and the domain of R'(x), which is also 0 ≤ x ≤ 100. The graph of R'(x) will represent the slope or the rate of change of the revenue function R(x) at each point.
Now, to find the number of units sold at which the marginal revenue begins to increase, we need to look for the x-values where the derivative R'(x) is greater than zero.
We already found the derivative R'(x) to be:
R'(x) = 90x^2 - 240x
To find when R'(x) > 0, we solve the inequality:
90x^2 - 240x > 0
Factor out common terms:
90x(x - 8/3) > 0
To find the critical points, we set each factor equal to zero:
x = 0 or x - 8/3 = 0
x = 8/3 or x = 0
We can now analyze the intervals using a sign chart or by considering test points in each interval:
For x < 0, we choose a test point x = -1:
90(-1)(-1 - 8/3) > 0
90/3(8/3 + 1) > 0
-30 > 0
Since -30 is less than zero, this interval does not satisfy the inequality R'(x) > 0.
For 0 < x < 8/3, we choose a test point x = 1:
90(1)(1 - 8/3) > 0
90/3(3/3 - 8/3) > 0
90/3(-5/3) > 0
-150/3 > 0
-50 > 0
Since -50 is less than zero, this interval also does not satisfy the inequality R'(x) > 0.
For 8/3 < x < 100, we choose a test point x = 10:
90(10)(10 - 8/3) > 0
90/3(30/3 - 8/3) > 0
90/3(22/3) > 0
660/3 > 0
220 > 0
Since 220 is greater than zero, this interval satisfies the inequality R'(x) > 0.
Therefore, the marginal revenue begins to increase at x = 8/3 or approximately 2.67 units sold.