2. The monthly revenue, R, for a big tech company depends on the monthly advertising expenses, x, by the
function r(x)=1020x / 25+8, where the monthly revenue and advertising expenses are in thousands of dollars.
a) What should be the monthly advertising expenses if the company wants the monthly revenue (in $) to
exceed $20000?
b) What will happen to the amount of monthly revenue as monthly advertising expenses increase? Provide a
complete answer by providing details about the behaviour of the graph and numerical details about the monthly
revenue
(a) just use your r(x) to solve r(x) > 20,000
(b) since the slope is positive, r(x) increases as x increases
a) We are given the equation for the monthly revenue as a function of advertising expenses:
R(x) = (1020x) / (25+8)
We want to find the monthly advertising expenses, x, when the monthly revenue exceeds $20,000.
We can set up the inequality:
R(x) > 20,000
Substituting the formula for R(x) and solving the inequality, we have:
(1020x) / (25+8) > 20,000
Multiplying both sides by (25+8) to eliminate the denominator, we get:
1020x > 20,000 * (25+8)
Simplifying further:
1020x > 26,000 * 20
Dividing both sides by 1020:
x > (26,000 * 20) / 1020
Simplifying:
x > 520,000 / 1020
x > 509.80392
Therefore, the monthly advertising expenses should be more than $509.80 (rounded to two decimal places) for the revenue to exceed $20,000.
b) As the monthly advertising expenses increase, the amount of monthly revenue generally increases as well.
This can be observed from the equation R(x) = (1020x) / (25+8).
If we examine the equation, we can see that "x" is in the numerator, while the denominator (25+8) is a constant.
This means that as "x" increases, the overall fraction increases, resulting in a higher monthly revenue.
Numerically, we can observe this behavior by plugging in different values for "x" and calculating the corresponding monthly revenue.
For example, if we assume x = 100, the monthly revenue would be:
R(100) = (1020 * 100) / (25 + 8) = 1050
If we increase "x" to 200:
R(200) = (1020 * 200) / (25 + 8) = 2100
We can see that doubling the advertising expenses results in doubling the monthly revenue.
Therefore, as the monthly advertising expenses increase, the amount of monthly revenue increases, showing a positive correlation between the two variables.
a) To find the monthly advertising expenses where the monthly revenue exceeds $20000, we can set up an inequality using the given function:
r(x) > 20000
Substituting the given function into the inequality:
1020x / (25+8) > 20000
Simplifying the expression:
1020x / 33 > 20000
Multiply both sides of the inequality by 33:
1020x > 20000 * 33
Divide both sides of the inequality by 1020:
x > (20000 * 33) / 1020
Calculating the result:
x > 660
Therefore, the monthly advertising expenses should be greater than $660,000.
b) To understand the behavior of the graph and how the monthly revenue changes as the monthly advertising expenses increase, we can analyze the given function:
r(x) = 1020x / (25+8)
Looking at the numerator, 1020x, we can see that as the monthly advertising expenses (x) increase, the monthly revenue (r(x)) will also increase. This suggests a positive relationship between advertising expenses and revenue.
Now, examining the denominator, (25+8), as the value of x increases, the denominator only changes by a constant value of 8. This means that the denominator has a minimal impact on the overall value of r(x).
Considering these aspects, we can conclude that as the monthly advertising expenses increase, the amount of monthly revenue will also increase. The rate of increase will depend on the specific value of x and can be obtained by evaluating the function for different values of x.