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 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 plot the points on a coordinate plane.
a. Graphing R(x):
Since R(x) is a cubic function, we know it will have a general shape resembling a curve. To sketch the graph of R(x), you can plot a few key points and then draw a smooth curve passing through those points.
Let's choose x-values of 0, 50, and 100 for simplicity. Plug these values into the equation R(x) = 30x^3 - 120x^2 + 500 to find the corresponding y-values:
R(0) = 30(0)^3 - 120(0)^2 + 500 = 500
R(50) = 30(50)^3 - 120(50)^2 + 500 = 375,000
R(100) = 30(100)^3 - 120(100)^2 + 500 = 2,900,500
Now, plot the points (0, 500), (50, 375,000), and (100, 2,900,500) on the coordinate plane. Connect these points with a smooth curve.
b. Graphing R'(x):
To graph R'(x), we need to find the derivative of R(x) using the power rule. The derivative of each term can be calculated individually, so let's find the derivatives of the three terms separately:
d/dx (30x^3) = 90x^2
d/dx (-120x^2) = -240x
d/dx (500) = 0 (since it's a constant)
Now we can combine these derivatives to find R'(x):
R'(x) = 90x^2 - 240x
To find the x-values at which the marginal revenue begins to increase, we need to find the critical points. Critical points occur when the derivative is equal to zero or undefined.
Set R'(x) = 0 and solve for x:
90x^2 - 240x = 0
Factor out 30x:
30x(3x - 8) = 0
Setting each factor equal to zero:
30x = 0 (gives x = 0 as a critical point)
3x - 8 = 0 (gives x = 8/3 or approximately 2.67 as another critical point)
So, the critical points are x = 0 and x = 8/3 (or 2.67).
The marginal revenue begins to increase at x = 8/3 (2.67 units sold).