The Revenue, In Dollars, Of A Company That Makes Toy Cars Can Be Modeled By The Polynomial 3 X 2 + 4 X − 60 3x^2 + 4x - 60 3 X 2 + 4 X − 60 . The Cost, In Dollars, Of Producing The Toy Cars Can Be Modeled By 3 X 2 − X + 200 3x^2 - X + 200 3 X 2 − X + 200 . The Number Of Toy Cars Sold Is

Introduction

In the world of business, understanding the revenue and cost of a product is crucial for making informed decisions. In this article, we will explore the revenue and cost of a company that manufactures toy cars using polynomial models. We will analyze the given polynomials and determine the number of toy cars sold.

Revenue and Cost Models

The revenue of the company can be modeled by the polynomial $3x^2 + 4x - 60$, where $x$ represents the number of toy cars sold. The cost of producing the toy cars can be modeled by the polynomial $3x^2 - x + 200$. These polynomials provide a mathematical representation of the revenue and cost of the company.

Revenue Model: $3x^2 + 4x - 60$

The revenue model is a quadratic polynomial, which means it has a parabolic shape. The coefficient of the $x^2$ term is 3, indicating that the revenue increases at a rate of 3 dollars per unit of $x$. The coefficient of the $x$ term is 4, indicating that the revenue increases by 4 dollars for every additional unit of $x$. The constant term is -60, indicating that the revenue is initially negative.

Cost Model: $3x^2 - x + 200$

The cost model is also a quadratic polynomial, with a parabolic shape. The coefficient of the $x^2$ term is 3, indicating that the cost increases at a rate of 3 dollars per unit of $x$. The coefficient of the $x$ term is -1, indicating that the cost decreases by 1 dollar for every additional unit of $x$. The constant term is 200, indicating that the cost is initially positive.

Profit Model

The profit of the company is the difference between the revenue and the cost. To find the profit model, we subtract the cost model from the revenue model:

$Profit = Revenue - Cost$ $Profit = (3x^2 + 4x - 60) - (3x^2 - x + 200)$ $Profit = 5x - 260$

The profit model is a linear polynomial, indicating that the profit increases at a constant rate of 5 dollars per unit of $x$.

Number of Toy Cars Sold

To find the number of toy cars sold, we need to find the value of $x$ that maximizes the profit. Since the profit model is a linear polynomial, the maximum profit occurs when $x$ is at its maximum value. To find the maximum value of $x$, we need to find the vertex of the profit model.

Vertex of the Profit Model

The vertex of a linear polynomial is its maximum or minimum point. To find the vertex of the profit model, we need to find the value of $x$ that makes the derivative of the profit model equal to zero.

Let's find the derivative of the profit model:

$Profit = 5x - 260$ $Profit' = 5$

Since the derivative is a constant, the profit model is a linear polynomial, and its vertex is its maximum point. The maximum value of $x$ occurs when the profit is at its maximum value.

Maximum Profit

To find the maximum profit, we need to find the value of $x$ that makes the profit model equal to its maximum value. Since the profit model is a linear polynomial, its maximum value occurs when $x$ is at its maximum value.

Let's find the maximum value of $x$:

$Profit = 5x - 260$ $Profit_{max} = 5x_{max} - 260$

To find the maximum value of $x$, we need to find the value of $x$ that makes the profit model equal to its maximum value. Since the profit model is a linear polynomial, its maximum value occurs when $x$ is at its maximum value.

Let's find the maximum value of $x$:

$x_{max} = \frac{260}{5}$ $x_{max} = 52$

The maximum value of $x$ is 52, which means that the company sells 52 toy cars to maximize its profit.

Conclusion

In this article, we analyzed the revenue and cost of a company that manufactures toy cars using polynomial models. We found the profit model and determined the number of toy cars sold to maximize the profit. The maximum value of $x$ is 52, which means that the company sells 52 toy cars to maximize its profit.

References

• [1] "Polynomial Models" by Math Is Fun
• [2] "Quadratic Polynomials" by Khan Academy
• [3] "Linear Polynomials" by Wolfram MathWorld

Appendix

The following is a list of the polynomials used in this article:

• Revenue model: $3x^2 + 4x - 60$
• Cost model: $3x^2 - x + 200$
• Profit model: $5x - 260$

The following is a list of the derivatives used in this article:

• Derivative of the revenue model: $6x + 4$
• Derivative of the cost model: $6x - 1$
• Derivative of the profit model: $5$
  The Revenue and Cost of Toy Cars: A Mathematical Analysis - Q&A ===========================================================

Introduction

In our previous article, we analyzed the revenue and cost of a company that manufactures toy cars using polynomial models. We found the profit model and determined the number of toy cars sold to maximize the profit. In this article, we will answer some of the most frequently asked questions about the revenue and cost of toy cars.

Q&A

Q: What is the revenue model of the company?

A: The revenue model of the company is given by the polynomial $3x^2 + 4x - 60$, where $x$ represents the number of toy cars sold.

Q: What is the cost model of the company?

A: The cost model of the company is given by the polynomial $3x^2 - x + 200$, where $x$ represents the number of toy cars sold.

Q: How do you find the profit model?

A: To find the profit model, we subtract the cost model from the revenue model:

$Profit = Revenue - Cost$ $Profit = (3x^2 + 4x - 60) - (3x^2 - x + 200)$ $Profit = 5x - 260$

Q: What is the maximum value of $x$?

A: The maximum value of $x$ is 52, which means that the company sells 52 toy cars to maximize its profit.

Q: How do you find the vertex of the profit model?

A: To find the vertex of the profit model, we need to find the value of $x$ that makes the derivative of the profit model equal to zero. Since the derivative is a constant, the profit model is a linear polynomial, and its vertex is its maximum point.

Q: What is the derivative of the profit model?

A: The derivative of the profit model is 5.

Q: How do you find the maximum profit?

A: To find the maximum profit, we need to find the value of $x$ that makes the profit model equal to its maximum value. Since the profit model is a linear polynomial, its maximum value occurs when $x$ is at its maximum value.

Q: What is the maximum profit?

A: The maximum profit is $5x_{max} - 260$, where $x_{max}$ is the maximum value of $x$. Since $x_{max}$ is 52, the maximum profit is $5(52) - 260 = 260 - 260 = 0$.

Q: Why is the maximum profit 0?

A: The maximum profit is 0 because the profit model is a linear polynomial, and its maximum value occurs when $x$ is at its maximum value. In this case, the maximum value of $x$ is 52, which means that the company sells 52 toy cars to maximize its profit. However, the profit model is a linear polynomial, and its maximum value occurs when $x$ is at its maximum value, which means that the maximum profit is 0.

Conclusion

In this article, we answered some of the most frequently asked questions about the revenue and cost of toy cars. We hope that this article has provided you with a better understanding of the revenue and cost of toy cars.

References

• [1] "Polynomial Models" by Math Is Fun
• [2] "Quadratic Polynomials" by Khan Academy
• [3] "Linear Polynomials" by Wolfram MathWorld

Appendix

The following is a list of the polynomials used in this article:

• Revenue model: $3x^2 + 4x - 60$
• Cost model: $3x^2 - x + 200$
• Profit model: $5x - 260$

The following is a list of the derivatives used in this article:

• Derivative of the revenue model: $6x + 4$
• Derivative of the cost model: $6x - 1$
• Derivative of the profit model: $5$