Select The Correct Answer.An Entrepreneur Estimates His Total Profit (total Revenue Minus Total Cost) For His Proposed Company As $p(x) = X^3 - 4x^2 + 5x - 20$, Where $p$ Is In Hundreds Of Dollars And $x$ Is The Number Of

As an entrepreneur, estimating the total profit of a proposed company is a crucial step in determining its potential for success. In this article, we will explore how to maximize profit using a mathematical approach, focusing on the given polynomial function $p(x) = x^3 - 4x^2 + 5x - 20$, where $p$ is in hundreds of dollars and $x$ is the number of units sold.

Understanding the Problem

The given polynomial function represents the total profit of the company, and our goal is to find the value of $x$ that maximizes this profit. To do this, we need to analyze the function and identify its critical points, which are the values of $x$ where the function changes from increasing to decreasing or vice versa.

Critical Points and the First Derivative

To find the critical points, we need to take the first derivative of the function, which represents the rate of change of the function with respect to $x$. The first derivative of $p(x)$ is given by:

$$p'(x) = 3x^2 - 8x + 5$$

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

$$3x^2 - 8x + 5 = 0$$

Using the quadratic formula, we get:

$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(5)}}{2(3)}$$

Simplifying the expression, we get:

$$x = \frac{8 \pm \sqrt{64 - 60}}{6}$$

$$x = \frac{8 \pm \sqrt{4}}{6}$$

$$x = \frac{8 \pm 2}{6}$$

Therefore, the critical points are:

$$x = \frac{8 + 2}{6} = \frac{10}{6} = \frac{5}{3}$$

$$x = \frac{8 - 2}{6} = \frac{6}{6} = 1$$

Second Derivative and Concavity

To determine the nature of the critical points, we need to take the second derivative of the function, which represents the rate of change of the first derivative with respect to $x$. The second derivative of $p(x)$ is given by:

$$p''(x) = 6x - 8$$

To determine the concavity of the function, we need to evaluate the second derivative at the critical points. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down.

Evaluating the second derivative at $x = \frac{5}{3}$, we get:

$$p''\left(\frac{5}{3}\right) = 6\left(\frac{5}{3}\right) - 8 = 10 - 8 = 2$$

Since the second derivative is positive, the function is concave up at $x = \frac{5}{3}$.

Evaluating the second derivative at $x = 1$, we get:

$$p''(1) = 6(1) - 8 = 6 - 8 = -2$$

Since the second derivative is negative, the function is concave down at $x = 1$.

Maximizing Profit

Based on the analysis of the critical points and the concavity of the function, we can conclude that the maximum profit occurs at $x = \frac{5}{3}$.

To find the maximum profit, we need to evaluate the function at $x = \frac{5}{3}$:

$$p\left(\frac{5}{3}\right) = \left(\frac{5}{3}\right)^3 - 4\left(\frac{5}{3}\right)^2 + 5\left(\frac{5}{3}\right) - 20$$

Simplifying the expression, we get:

$$p\left(\frac{5}{3}\right) = \frac{125}{27} - \frac{100}{9} + \frac{25}{3} - 20$$

$$p\left(\frac{5}{3}\right) = \frac{125 - 300 + 225 - 540}{27}$$

$$p\left(\frac{5}{3}\right) = \frac{-590}{27}$$

Therefore, the maximum profit is $-\frac{590}{27}$ hundred dollars, which occurs when $x = \frac{5}{3}$ units are sold.

Conclusion

In this article, we used a mathematical approach to maximize the profit of a proposed company. We analyzed the given polynomial function, identified its critical points, and determined the concavity of the function. Based on this analysis, we concluded that the maximum profit occurs at $x = \frac{5}{3}$ units sold, resulting in a maximum profit of $-\frac{590}{27}$ hundred dollars.

References

• [1] Calculus: Early Transcendentals, 8th edition, James Stewart
• [2] Calculus, 3rd edition, Michael Spivak

Appendix

The following is a list of formulas and theorems used in this article:

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Second derivative test: If $p''(x) > 0$, the function is concave up; if $p''(x) < 0$, the function is concave down.
  Maximizing Profit: A Mathematical Approach to Entrepreneurial Success - Q&A ====================================================================

In our previous article, we explored how to maximize profit using a mathematical approach, focusing on the given polynomial function $p(x) = x^3 - 4x^2 + 5x - 20$, where $p$ is in hundreds of dollars and $x$ is the number of units sold. In this article, we will answer some of the most frequently asked questions related to maximizing profit.

Q: What is the main goal of maximizing profit?

A: The main goal of maximizing profit is to determine the optimal number of units to sell in order to maximize the total revenue of the company.

Q: How do I determine the optimal number of units to sell?

A: To determine the optimal number of units to sell, you need to analyze the given polynomial function and identify its critical points. The critical points are the values of $x$ where the function changes from increasing to decreasing or vice versa.

Q: What is the significance of the first derivative in maximizing profit?

A: The first derivative of the function represents the rate of change of the function with respect to $x$. It helps us identify the critical points and determine the concavity of the function.

Q: What is the significance of the second derivative in maximizing profit?

A: The second derivative of the function represents the rate of change of the first derivative with respect to $x$. It helps us determine the concavity of the function and identify the maximum or minimum points.

Q: How do I determine the concavity of the function?

A: To determine the concavity of the function, you need to evaluate the second derivative at the critical points. If the second derivative is positive, the function is concave up; if it is negative, the function is concave down.

Q: What is the maximum profit, and how do I find it?

A: The maximum profit is the highest value of the function, which occurs at the critical point where the function changes from increasing to decreasing. To find the maximum profit, you need to evaluate the function at the critical point.

Q: What are the limitations of using a mathematical approach to maximize profit?

A: One of the limitations of using a mathematical approach to maximize profit is that it assumes that the function is continuous and differentiable. In reality, the function may not be continuous or differentiable, which can affect the accuracy of the results.

Q: Can I use a mathematical approach to maximize profit in real-world scenarios?

A: Yes, you can use a mathematical approach to maximize profit in real-world scenarios, but you need to consider the limitations and assumptions of the approach. You also need to consider other factors that can affect the profit, such as market trends, competition, and economic conditions.

Q: What are some common mistakes to avoid when using a mathematical approach to maximize profit?

A: Some common mistakes to avoid when using a mathematical approach to maximize profit include:

• Assuming that the function is continuous and differentiable when it is not
• Ignoring the limitations and assumptions of the approach
• Failing to consider other factors that can affect the profit
• Not evaluating the function at the critical points

Conclusion

In this article, we answered some of the most frequently asked questions related to maximizing profit using a mathematical approach. We hope that this article has provided you with a better understanding of the topic and has helped you to avoid common mistakes. Remember to always consider the limitations and assumptions of the approach and to evaluate the function at the critical points to ensure accurate results.

References

• [1] Calculus: Early Transcendentals, 8th edition, James Stewart
• [2] Calculus, 3rd edition, Michael Spivak

Appendix

The following is a list of formulas and theorems used in this article:

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Second derivative test: If $p''(x) > 0$, the function is concave up; if $p''(x) < 0$, the function is concave down.