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The question asks for solving various parts related to the demand equation, which is given as:

\[
x = -5p + 200
\]

Where:
- \( p \) is the price in dollars.
- \( x \) is the quantity sold of a certain product.

Let's analyze each part of the problem:

### (a) Find a model that expresses the revenue \( R \) as a function of \( p \).
We know that revenue \( R \) is given by the formula:

\[
R = p \times x
\]

Substitute \( x = -5p + 200 \) into this equation:

\[
R(p) = p \times (-5p + 200)
\]

Simplify this:

\[
R(p) = -5p^2 + 200p
\]

This is the revenue function in terms of \( p \).

### (b) What is the domain of \( R \), assuming that \( R \geq 0 \)?

To find the domain, we ensure the revenue is nonnegative, i.e., \( R(p) \geq 0 \).

We already have:

\[
R(p) = -5p^2 + 200p
\]

This is a quadratic equation, and the revenue will be non-negative between the roots of the equation:

\[
-5p^2 + 200p = 0
\]

Factor this equation:

\[
p(-5p + 200) = 0
\]

So, \( p = 0 \) or \( p = 40 \).

Therefore, the domain is \( 0 \leq p \leq 40 \).

### (c) What price \( p \) maximizes revenue?

Since the revenue function is quadratic and opens downward (as the coefficient of \( p^2 \) is negative), the price that maximizes revenue is at the vertex of the parabola. The formula for the vertex is:

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

For \( R(p) = -5p^2 + 200p \), \( a = -5 \) and \( b = 200 \). Substituting into the vertex formula:

\[
p = -\frac{200}{2(-5)} = 20
\]

Thus, the price that maximizes revenue is \( p = 20 \).

### (d) What is the maximum revenue?

To find the maximum revenue, substitute \( p = 20 \) into the revenue function:

\[
R(20) = -5(20)^2 + 200(20)
\]
\[
R(20) = -5(400) + 4000 = -2000 + 4000 = 2000
\]

The maximum revenue is \( R = 2000 \).

### (e) How many units are sold at this price?

We use the demand equation \( x = -5p + 200 \) and substitute \( p = 20 \):

\[
x = -5(20) + 200 = -100 + 200 = 100
\]

So, 100 units are sold at this price.

### (f) Graph \( R \) and choose the correct graph.

The revenue function \( R(p) = -5p^2 + 200p \) is a downward-opening parabola. Based on the calculated maximum at \( p = 20 \) and \( R = 2000 \), the correct graph is **D**.

Would you like more details on any part of this solution?

Here are five related questions:
1. How do you calculate the roots of a quadratic function?
2. What happens to the revenue if the price increases beyond the maximum?
3. How can you find the vertex of a parabola algebraically?
4. How do we interpret the negative coefficient of \( p^2 \) in the revenue function?
5. What other factors can affect the domain of a revenue function?

**Tip:** When solving quadratic functions, always check if the parabola opens upwards or downwards, as this determines whether the vertex is a maximum or minimum.

The price p (in dollars) and the quantity x sold of a certain product satisfy the demand equation x = -5p + 200. Answer parts (a) through (g).

Learn how to express revenue as a function of price, find the domain, and determine the price that maximizes revenue using a quadratic function. This problem includes key concepts such as revenue function, demand equation, and finding the maximum value of a quadratic equation. Suitable for Grades 9-11 students.

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