The price p​ (in dollars) and the quantity x sold of a certain product satisfy the demand equation x equals negative 5 p plus 100. Answer parts​ (a) through​ (g).
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Part 1
​(a) Find a model that expresses the revenue R as a function of p.​ (Remember, Requals​xp.)
​R(p)equals

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To find a model that expresses the revenue \( R \) as a function of price \( p \), we need to use the relationship:

\[
R = x \cdot p
\]

We are also given the demand equation:

\[
x = -5p + 100
\]

Now, substitute this expression for \( x \) into the revenue equation:

\[
R = (-5p + 100) \cdot p
\]

Simplifying this expression:

\[
R(p) = -5p^2 + 100p
\]

Thus, the revenue \( R \) as a function of \( p \) is:

\[
R(p) = -5p^2 + 100p
\]

Would you like further details on this? 

Here are some related questions to consider:
1. What is the maximum revenue that can be achieved?
2. For what price \( p \) will the revenue be maximized?
3. How does the revenue change as the price increases?
4. How does the demand equation relate to real-world sales models?
5. What is the revenue when \( p = 10 \)?

**Tip:** The maximum revenue can be found by analyzing the vertex of the quadratic function for revenue.

The price p (in dollars) and the quantity x sold of a certain product satisfy the demand equation x = -5p + 100. Answer parts (a) through (g). Part (a): Find a model that expresses the revenue R as a function of p. (Remember, R = xp.)

Explore how to derive a revenue model from a given demand equation using algebraic principles. Learn to express revenue as a quadratic function and analyze the maximum revenue point. Suitable for high school students in Grades 9-11.

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