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The problem you're working on involves determining when a company's revenue will be zero and finding the domain for the revenue function.

### Problem Information:
The revenue function is given as:

\[
R(p) = 1000p(30 - p)
\]

### (a) **When the company has no revenue:**

The revenue will be zero when \( R(p) = 0 \). Thus, we need to solve the equation:

\[
1000p(30 - p) = 0
\]

This gives us two cases:
1. \( p = 0 \)
2. \( 30 - p = 0 \) → \( p = 30 \)

So, the company has no revenue when \( p = 0 \) or \( p = 30 \). The values should be entered in increasing order, so \( p = 0 \) and \( p = 30 \).

### (b) **Reasonable Domain for \( p \):**

Since revenue depends on the price \( p \), a reasonable domain would be where \( 0 \leq p \leq 30 \). This is because outside of this range, the revenue would either be negative or zero, which doesn't make sense for this scenario.

Let me know if you need more details, and here are some related questions to deepen your understanding:

### Questions:
1. What is the maximum revenue that the company can achieve?
2. How do you find the value of \( p \) that maximizes the revenue?
3. What is the significance of the values \( p = 0 \) and \( p = 30 \) in real-world terms?
4. How would the graph of the revenue function look like, and what does its shape tell us?
5. How does the company's revenue change as \( p \) increases from 0 to 30?

### Tip:
Always double-check the units of variables like price and revenue in word problems to ensure consistency in the solution.

When a company charges a price p dollars for one of its products, its revenue is given by R(p) = 1000p(30 - p). For what prices does the company have no revenue? What is a reasonable domain for p?

Learn how to determine the prices at which a company has no revenue using the quadratic revenue function R(p) = 1000p(30 - p). This problem covers key algebraic concepts like solving quadratic equations and defining a reasonable domain for the price variable. Suitable for students in Grades 9-12.

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