## Revenue Maximization Exercise ### Problem Statement Suppose that the selling price \( p \) of an item for the quantity \( x \) sold is given by the function: \[ p = -\frac{1}{10}x + 28 \] ### Questions 1. **Express the revenue \( R \) as a function of \( x \).** \[ R = x \cdot p \] \[ R = \] 2. **How many items must be sold to maximize the revenue?** \[ x = \] 3. **What is the maximum revenue that can be obtained from this model?** \[ \$ \] --- ### Explanation In this exercise, we are given a linear price function and asked to maximize revenue based on this function. The given price function is \( p = -\frac{1}{10}x + 28 \). The revenue \( R \) is calculated as the product of the quantity of items sold \( x \) and the price \( p \), i.e., \( R = x \cdot p \). Therefore, we need to: 1. Derive the revenue function \( R \) in terms of \( x \). 2. Determine the value of \( x \) that maximizes this revenue function. 3. Calculate the maximum possible revenue. Let's solve these steps systematically: 1. **Express the revenue \( R \) as a function of \( x \)**: \[ R = x \cdot \left(-\frac{1}{10}x + 28\right) \] \[ R = -\frac{1}{10}x^2 + 28x \] \[ R = \] 2. **How many items must be sold to maximize the revenue?** To find the quantity \( x \) that maximizes the quadratic revenue function \( R = -\frac{1}{10}x^2 + 28x \), we can use the vertex formula for a parabola \( x = -\frac{b}{2a} \), where \( a = -\frac{1}{10} \) and \( b = 28 \): \[ x = -\frac{28}{2 \cdot -\frac{1}{10}} \] \[ x = 140 \] \[ x = \] 3. **What is the maximum revenue that can be obtained from this model?** Substitute \( x =
## Revenue Maximization Exercise ### Problem Statement Suppose that the selling price \( p \) of an item for the quantity \( x \) sold is given by the function: \[ p = -\frac{1}{10}x + 28 \] ### Questions 1. **Express the revenue \( R \) as a function of \( x \).** \[ R = x \cdot p \] \[ R = \] 2. **How many items must be sold to maximize the revenue?** \[ x = \] 3. **What is the maximum revenue that can be obtained from this model?** \[ \$ \] --- ### Explanation In this exercise, we are given a linear price function and asked to maximize revenue based on this function. The given price function is \( p = -\frac{1}{10}x + 28 \). The revenue \( R \) is calculated as the product of the quantity of items sold \( x \) and the price \( p \), i.e., \( R = x \cdot p \). Therefore, we need to: 1. Derive the revenue function \( R \) in terms of \( x \). 2. Determine the value of \( x \) that maximizes this revenue function. 3. Calculate the maximum possible revenue. Let's solve these steps systematically: 1. **Express the revenue \( R \) as a function of \( x \)**: \[ R = x \cdot \left(-\frac{1}{10}x + 28\right) \] \[ R = -\frac{1}{10}x^2 + 28x \] \[ R = \] 2. **How many items must be sold to maximize the revenue?** To find the quantity \( x \) that maximizes the quadratic revenue function \( R = -\frac{1}{10}x^2 + 28x \), we can use the vertex formula for a parabola \( x = -\frac{b}{2a} \), where \( a = -\frac{1}{10} \) and \( b = 28 \): \[ x = -\frac{28}{2 \cdot -\frac{1}{10}} \] \[ x = 140 \] \[ x = \] 3. **What is the maximum revenue that can be obtained from this model?** Substitute \( x =
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![## Revenue Maximization Exercise
### Problem Statement
Suppose that the selling price \( p \) of an item for the quantity \( x \) sold is given by the function:
\[ p = -\frac{1}{10}x + 28 \]
### Questions
1. **Express the revenue \( R \) as a function of \( x \).**
\[ R = x \cdot p \]
\[ R = \]
2. **How many items must be sold to maximize the revenue?**
\[ x = \]
3. **What is the maximum revenue that can be obtained from this model?**
\[ \$ \]
---
### Explanation
In this exercise, we are given a linear price function and asked to maximize revenue based on this function. The given price function is \( p = -\frac{1}{10}x + 28 \).
The revenue \( R \) is calculated as the product of the quantity of items sold \( x \) and the price \( p \), i.e., \( R = x \cdot p \). Therefore, we need to:
1. Derive the revenue function \( R \) in terms of \( x \).
2. Determine the value of \( x \) that maximizes this revenue function.
3. Calculate the maximum possible revenue.
Let's solve these steps systematically:
1. **Express the revenue \( R \) as a function of \( x \)**:
\[ R = x \cdot \left(-\frac{1}{10}x + 28\right) \]
\[ R = -\frac{1}{10}x^2 + 28x \]
\[ R = \]
2. **How many items must be sold to maximize the revenue?**
To find the quantity \( x \) that maximizes the quadratic revenue function \( R = -\frac{1}{10}x^2 + 28x \), we can use the vertex formula for a parabola \( x = -\frac{b}{2a} \), where \( a = -\frac{1}{10} \) and \( b = 28 \):
\[ x = -\frac{28}{2 \cdot -\frac{1}{10}} \]
\[ x = 140 \]
\[ x = \]
3. **What is the maximum revenue that can be obtained from this model?**
Substitute \( x =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2c5c293-7ff8-490b-b548-13e2f43ca7a8%2Ffe7ce0e0-4486-4b22-bd1e-a8fa158404dc%2F3zk7res.jpeg&w=3840&q=75)
Transcribed Image Text:## Revenue Maximization Exercise
### Problem Statement
Suppose that the selling price \( p \) of an item for the quantity \( x \) sold is given by the function:
\[ p = -\frac{1}{10}x + 28 \]
### Questions
1. **Express the revenue \( R \) as a function of \( x \).**
\[ R = x \cdot p \]
\[ R = \]
2. **How many items must be sold to maximize the revenue?**
\[ x = \]
3. **What is the maximum revenue that can be obtained from this model?**
\[ \$ \]
---
### Explanation
In this exercise, we are given a linear price function and asked to maximize revenue based on this function. The given price function is \( p = -\frac{1}{10}x + 28 \).
The revenue \( R \) is calculated as the product of the quantity of items sold \( x \) and the price \( p \), i.e., \( R = x \cdot p \). Therefore, we need to:
1. Derive the revenue function \( R \) in terms of \( x \).
2. Determine the value of \( x \) that maximizes this revenue function.
3. Calculate the maximum possible revenue.
Let's solve these steps systematically:
1. **Express the revenue \( R \) as a function of \( x \)**:
\[ R = x \cdot \left(-\frac{1}{10}x + 28\right) \]
\[ R = -\frac{1}{10}x^2 + 28x \]
\[ R = \]
2. **How many items must be sold to maximize the revenue?**
To find the quantity \( x \) that maximizes the quadratic revenue function \( R = -\frac{1}{10}x^2 + 28x \), we can use the vertex formula for a parabola \( x = -\frac{b}{2a} \), where \( a = -\frac{1}{10} \) and \( b = 28 \):
\[ x = -\frac{28}{2 \cdot -\frac{1}{10}} \]
\[ x = 140 \]
\[ x = \]
3. **What is the maximum revenue that can be obtained from this model?**
Substitute \( x =
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