Suppose that the manufacturer of a gas clothes dryer has found​ that, when the unit price is p​ dollars, the revenue R​ (in dollars) is Upper R left parenthesis p right parenthesis equals negative 5 p squared plus 10 comma 000 p. What unit price should be established for the dryer to maximize​ revenue? What is the maximum​ revenue?
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Part 1
The unit price that should be established to maximize revenue is ​$

enter your response here.

To solve for the unit price that maximizes revenue, we need to find the price $ p $ that maximizes the revenue function:

\[
R(p) = -5p^2 + 10,000p
\]

This is a quadratic function, and the maximum of a quadratic function $ ax^2 + bx + c $ occurs at:

\[
p = -\frac{b}{2a}
\]

For our equation, $ a = -5 $ and $ b = 10,000 $. Plugging these values into the formula:

\[
p = -\frac{10,000}{2(-5)} = \frac{10,000}{10} = 1,000
\]

Thus, the unit price that should be established to maximize revenue is **$1,000**.

---

### Part 2: Maximum Revenue

To find the maximum revenue, substitute $ p = 1,000 $ back into the revenue function:

\[
R(1,000) = -5(1,000)^2 + 10,000(1,000)
\]
\[
R(1,000) = -5(1,000,000) + 10,000,000
\]
\[
R(1,000) = -5,000,000 + 10,000,000 = 5,000,000
\]

Thus, the maximum revenue is **$5,000,000**.

---

### Summary:
- The unit price that should be established to maximize revenue is **$1,000**.
- The maximum revenue is **$5,000,000**.

---

Do you have any questions or want further details?

### Related Questions:
1. How do you find the maximum of a general quadratic function?
2. What happens to revenue if the price is increased above $1,000?
3. Can a quadratic function have more than one maximum point?
4. What are real-world factors that could affect the revenue aside from price?
5. How do changes in production cost affect the revenue model?

### Tip:
Always check the sign of the quadratic term in $ ax^2 $. If $ a $ is negative, the parabola opens downwards, meaning it has a maximum point.

Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p​ dollars, the revenue R​ (in dollars) is R(p) = -5p^2 + 10,000p. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?

Learn how to maximize revenue for a gas clothes dryer using quadratic functions. The revenue function R(p) = -5p^2 + 10,000p is solved by applying the vertex formula to find the optimal unit price of $1,000 and maximum revenue of $5,000,000. Suitable for high school students in Grades 10-12.

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