(a) Weekly revenue when the price is set at $20 per ruby:

The revenue is approximately $2,147.

(b) Weekly revenue when the price is set at $200 per ruby:

The revenue is approximately -$62,533.

Interpretation:

The negative revenue when the price is set at $200 suggests that at this price point, the model predicts a significant loss for Royal Ruby Retailers. This implies that setting such a high price is not realistic within the function's practical constraints.

• Domain of the function likely includes $p = 200$, but charging this price leads to a revenue loss.
• Range of the function cannot realistically include such a large negative revenue at high prices, suggesting the demand decreases significantly.

Now, I'll proceed to part (c), determining the price that would result in a weekly revenue of $2,541.

It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like me to guide you through the process of solving part (c) manually or assist with anything else?

Here are five questions that could expand on this:

1. How does the price affect the revenue in general based on the given function?
2. What does the domain of a function represent in real-world terms for businesses?
3. Could you graph the revenue function to visualize how price changes affect revenue?
4. How can Royal Ruby Retailers determine the best price to maximize revenue using calculus?
5. What is the significance of negative revenue in this context?

Tip: When solving quadratic equations like this, it's useful to check for both solutions, as one may not be practical for the real-world scenario.