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The problem provided is asking for Apollo's revenue when 5,000 units are produced, using the revenue function:

\[
R(x) = -0.1x^2 + 800x
\]

Where $ x $ represents the number of units manufactured. We are tasked with finding $ R(5000) $, so we can substitute $ x = 5000 $ into the function and calculate the revenue.

\[
R(5000) = -0.1(5000)^2 + 800(5000)
\]

Let's calculate this step-by-step:

1. $ (5000)^2 = 25,000,000 $
2. $ -0.1 \times 25,000,000 = -2,500,000 $
3. $ 800 \times 5000 = 4,000,000 $
4. $ R(5000) = -2,500,000 + 4,000,000 = 1,500,000 $

Thus, Apollo's revenue when 5,000 units are produced is **$1,500,000**.

Would you like more details or have any questions about this? Here are 5 related questions to expand on this:

1. How does the revenue function change if the production cost increases by 10%?
2. What would the revenue be if 10,000 units are produced?
3. How can we determine the maximum possible revenue using this quadratic function?
4. What is the break-even point where revenue equals zero?
5. How does the graph of the revenue function look and how does it help in understanding the revenue trend?

**Tip**: In revenue functions of the form $ R(x) = ax^2 + bx $, the term with $ x^2 $ usually represents diminishing returns after a certain production level.

The revenue (in dollars) realized by Apollo from the sale of its ink-jet printers is given by R(x) = -0.1x^2 + 800x where x denotes the number of units manufactured each month. What is Apollo's revenue (in dollars) when 5,000 units are produced?

This problem explores a quadratic revenue function R(x) = -0.1x^2 + 800x. Learn how to calculate Apollo's revenue when 5,000 units are produced, applying algebraic and quadratic function concepts. Ideal for high school students in grades 10-12.

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