Advanced Polynomial & Rational Functions
The monthly revenue,R(x),for a big tech company depends on the monthly advertising expenses, x, by the function 𝑅(𝑥) = 1020𝑥/25 + 8x , 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 dollars ) to exceed $200 000? Include an interval chart in your solution.

1 Answer
Note that
\[𝑅(𝑥) = \frac{1020 x}{25} + 8x=40.8x+8x=48.8 x\]
a) If the company wants the monthly revenue (in dollars ) to exceed 200 000 dollars, then we should have
\[𝑅(𝑥) = 48.8 x \geq 200 \Rightarrow x \geq \frac{200}{48.8}=4.098.\]
So the monthly advertising expenses should be at least $4.098 k=4,098$ dollars. So the monthly advertising expenses should be a number in the interval $(4,098, \infty).$

- 1 Answer
- 371 views
- Pro Bono
Related Questions
- Is it true almost all Lebesgue measurable functions are non-integrable?
- Compounding interest of principal P, where a compounding withdrawal amount W get withdrawn from P before each compounding of P.
- Mechanical principle help (maths)
- Using Substitution to Prove an Big O/upper bound is O(n^3)
- Can someone translate $s_j : \Omega \hspace{3pt} x \hspace{3pt} [0,T_{Final}] \rightarrow S_j \subset R$ into simple English for me?
- What can we say about the transformation ?
- Studying the graph of this function
- Amendment to "Same exponents in two nonlinear function deriving the same result?"
Could you clarify what you mean by "the monthly revenue and advertising expenses are in 25+8𝑥 thousands of dollars."
made a mistake, its supposed to be R(x) = 1020x/25 + 8x
You question is not clear. Please check your question carefully and revise it.
Could you please revise your question?
i fixed it