The relationship between the amount of money x that Cannon Precision Instruments spends on advertising and the company's total sales S(x) is given by the following function where x is measured in thousands of dollars.
S(x) = -0.002x3 + 0.9x2 + 4x + 500 (0 x 200)
Find the rate of change of the sales with respect to the amount of money spent on advertising.
S'(x) =
Are Cannon's total sales increasing at a faster rate when the amount of money spent on advertising is $100,000 or $150,000?
Rick
To find the rate of change of the sales (S(x)) with respect to the amount of money spent on advertising (x), we need to find the derivative of the sales function (S(x)).
The derivative of a function represents the rate of change of that function. So, to find the rate of change in sales with respect to advertising spending, we need to find the derivative of the sales function.
Let's differentiate the sales function S(x) using the power rule of differentiation:
S(x) = -0.002x^3 + 0.9x^2 + 4x + 500
To differentiate, take the derivative of each term separately:
d/dx (-0.002x^3) = -0.006x^2
d/dx (0.9x^2) = 1.8x
d/dx (4x) = 4
As the derivative of a constant term is zero, the derivative of 500 is 0.
Therefore, the derivative of the sales function (S'(x)) is:
S'(x) = -0.006x^2 + 1.8x + 4
Now, to determine whether the total sales are increasing at a faster rate when the amount of money spent on advertising is $100,000 or $150,000, we need to compare the values of S'(x) at those points.
For $100,000 (x = 100), we can evaluate S'(x):
S'(100) = -0.006(100)^2 + 1.8(100) + 4
For $150,000 (x = 150), we can evaluate S'(x):
S'(150) = -0.006(150)^2 + 1.8(150) + 4
By comparing these values, we can determine if the rate of change of sales is higher at $100,000 or $150,000.