A Sock Company Estimates That Its Monthly Cost Is C ( X ) = 100 X C(x) = 100x C ( X ) = 100 X And Its Monthly Revenue Is R ( X ) = − 0.5 X 3 + 800 X 2 − 700 X R(x) = -0.5x^3 + 800x^2 - 700x R ( X ) = − 0.5 X 3 + 800 X 2 − 700 X , Where X X X Is In Thousands Of Pairs Of Socks Sold. The Profit Is The Difference Between The Revenue

Introduction

In the world of business, profit maximization is a crucial goal for companies to achieve. A sock company, like any other business, aims to maximize its profit by selling a certain number of pairs of socks. In this article, we will explore how a sock company can estimate its monthly cost and revenue, and ultimately, its profit. We will use mathematical functions to model the company's cost and revenue, and then find the profit function.

The Cost Function

The cost function, denoted by $C(x)$, represents the total cost incurred by the company in producing and selling $x$ thousands of pairs of socks. In this case, the cost function is given by:

$$C(x) = 100x$$

where $x$ is in thousands of pairs of socks sold. This means that for every thousand pairs of socks sold, the company incurs a cost of $100.

The Revenue Function

The revenue function, denoted by $R(x)$, represents the total revenue generated by the company from selling $x$ thousands of pairs of socks. In this case, the revenue function is given by:

$$R(x) = -0.5x^3 + 800x^2 - 700x$$

where $x$ is in thousands of pairs of socks sold. This means that the revenue generated by the company is a cubic function of the number of pairs of socks sold.

The Profit Function

The profit function, denoted by $P(x)$, represents the difference between the revenue and the cost incurred by the company. In this case, the profit function is given by:

$$P(x) = R(x) - C(x)$$

Substituting the expressions for $R(x)$ and $C(x)$, we get:

$$P(x) = (-0.5x^3 + 800x^2 - 700x) - 100x$$

Simplifying the expression, we get:

$$P(x) = -0.5x^3 + 800x^2 - 800x$$

Finding the Maximum Profit

To find the maximum profit, we need to find the critical points of the profit function. Critical points occur when the derivative of the function is equal to zero or undefined. In this case, we need to find the derivative of the profit function with respect to $x$.

Using the power rule of differentiation, we get:

$$P'(x) = -1.5x^2 + 1600x - 800$$

To find the critical points, we set the derivative equal to zero and solve for $x$:

$$-1.5x^2 + 1600x - 800 = 0$$

Solving this quadratic equation, we get:

$$x = \frac{-1600 \pm \sqrt{1600^2 - 4(-1.5)(-800)}}{2(-1.5)}$$

Simplifying the expression, we get:

$$x = \frac{-1600 \pm \sqrt{2560000 - 4800}}{-3}$$

$$x = \frac{-1600 \pm \sqrt{2555200}}{-3}$$

$$x = \frac{-1600 \pm 1599.999}{-3}$$

$$x = \frac{-1600 + 1599.999}{-3} \text{ or } x = \frac{-1600 - 1599.999}{-3}$$

$$x = \frac{-0.001}{-3} \text{ or } x = \frac{-3199.999}{-3}$$

$$x = 0.000333 \text{ or } x = 1066.666$$

Since $x$ represents the number of pairs of socks sold in thousands, we can ignore the negative value and consider only the positive value:

$$x = 1066.666$$

This means that the company should sell approximately 1067 thousand pairs of socks to maximize its profit.

Conclusion

In this article, we have estimated the monthly cost and revenue of a sock company using mathematical functions. We have also found the profit function and used it to determine the number of pairs of socks that the company should sell to maximize its profit. The company should sell approximately 1067 thousand pairs of socks to maximize its profit.

References

• [1] Cost and Revenue Functions. (n.d.). Retrieved from https://www.example.com/cost-and-revenue-functions
• [2] Profit Maximization. (n.d.). Retrieved from https://www.example.com/profit-maximization

Note

Introduction

In our previous article, we explored how a sock company can estimate its monthly cost and revenue, and ultimately, its profit. We used mathematical functions to model the company's cost and revenue, and then found the profit function. In this article, we will answer some frequently asked questions related to the sock company's profit maximization problem.

Q: What is the cost function, and how is it related to the number of pairs of socks sold?

A: The cost function, denoted by $C(x)$, represents the total cost incurred by the company in producing and selling $x$ thousands of pairs of socks. In this case, the cost function is given by:

$$C(x) = 100x$$

where $x$ is in thousands of pairs of socks sold. This means that for every thousand pairs of socks sold, the company incurs a cost of $100.

Q: What is the revenue function, and how is it related to the number of pairs of socks sold?

A: The revenue function, denoted by $R(x)$, represents the total revenue generated by the company from selling $x$ thousands of pairs of socks. In this case, the revenue function is given by:

$$R(x) = -0.5x^3 + 800x^2 - 700x$$

where $x$ is in thousands of pairs of socks sold. This means that the revenue generated by the company is a cubic function of the number of pairs of socks sold.

Q: How do we find the profit function, and what is its significance?

A: The profit function, denoted by $P(x)$, represents the difference between the revenue and the cost incurred by the company. In this case, the profit function is given by:

$$P(x) = R(x) - C(x)$$

Substituting the expressions for $R(x)$ and $C(x)$, we get:

$$P(x) = (-0.5x^3 + 800x^2 - 700x) - 100x$$

Simplifying the expression, we get:

$$P(x) = -0.5x^3 + 800x^2 - 800x$$

The profit function is significant because it helps us determine the number of pairs of socks that the company should sell to maximize its profit.

Q: How do we find the maximum profit, and what is the optimal number of pairs of socks to sell?

A: To find the maximum profit, we need to find the critical points of the profit function. Critical points occur when the derivative of the function is equal to zero or undefined. In this case, we need to find the derivative of the profit function with respect to $x$.

Using the power rule of differentiation, we get:

$$P'(x) = -1.5x^2 + 1600x - 800$$

To find the critical points, we set the derivative equal to zero and solve for $x$:

$$-1.5x^2 + 1600x - 800 = 0$$

Solving this quadratic equation, we get:

$$x = \frac{-1600 \pm \sqrt{1600^2 - 4(-1.5)(-800)}}{2(-1.5)}$$

Simplifying the expression, we get:

$$x = \frac{-1600 \pm \sqrt{2555200}}{-3}$$

$$x = \frac{-1600 \pm 1599.999}{-3}$$

$$x = \frac{-1600 + 1599.999}{-3} \text{ or } x = \frac{-1600 - 1599.999}{-3}$$

$$x = \frac{-0.001}{-3} \text{ or } x = \frac{-3199.999}{-3}$$

$$x = 0.000333 \text{ or } x = 1066.666$$

Since $x$ represents the number of pairs of socks sold in thousands, we can ignore the negative value and consider only the positive value:

$$x = 1066.666$$

This means that the company should sell approximately 1067 thousand pairs of socks to maximize its profit.

Q: What are some potential challenges or limitations of this approach?

A: Some potential challenges or limitations of this approach include:

• The cost and revenue functions may not accurately reflect the company's actual costs and revenues.
• The profit function may not be a good representation of the company's true profit.
• The company may face external factors such as changes in market demand, competition, or economic conditions that can affect its profit.
• The company may not have the resources or capabilities to produce and sell the optimal number of pairs of socks.

Conclusion

In this article, we have answered some frequently asked questions related to the sock company's profit maximization problem. We have discussed the cost and revenue functions, the profit function, and the optimal number of pairs of socks to sell. We have also highlighted some potential challenges or limitations of this approach.