https://file.aiquickdraw.com/mathbot/file/172947109713311nzevcd.jpg

The problem asks for a model that expresses the revenue \( R \) as a function of price \( p \), where:

- The demand equation is \( x = -9p + 900 \) (this gives the quantity \( x \) sold at price \( p \)).
- Revenue is defined as \( R = x \cdot p \).

### Step-by-step solution:

1. **Use the demand equation**:
   From the problem, we know that \( x = -9p + 900 \). 

2. **Substitute this into the revenue formula**:
   Since \( R = x \cdot p \), and \( x = -9p + 900 \), we substitute \( x \) into the revenue equation:
   \[
   R(p) = p \cdot (-9p + 900)
   \]

3. **Simplify the equation**:
   Now, distribute \( p \) on the right-hand side:
   \[
   R(p) = -9p^2 + 900p
   \]

Thus, the revenue function is:
\[
R(p) = -9p^2 + 900p
\]

This is the model that expresses revenue \( R \) as a function of price \( p \).

Let me know if you'd like more details or have questions.

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Here are 5 related questions to expand your understanding:

1. What does the coefficient of \( p^2 \) in the revenue function represent?
2. How does the price \( p \) affect the revenue, and when is revenue maximized?
3. Can you find the price \( p \) that maximizes revenue using this model?
4. What happens to revenue if the price is set too high or too low?
5. How would the model change if the demand equation had a different slope?

**Tip**: To maximize revenue, find the vertex of the quadratic function \( R(p) = -9p^2 + 900p \).

Find a model that expresses the revenue R as a function of p, where the demand equation is x = -9p + 900, and revenue is defined as R = xp.

Learn how to derive a revenue function R(p) from a demand equation x = -9p + 900. This step-by-step solution explains key algebraic concepts, using both linear and quadratic equations to model revenue. Suitable for students in Grades 9-11.

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