What Is The Maximum Value Of The Product $P = Xy$ When $y = 5 - 2x$?

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Introduction

In mathematics, optimization problems are a crucial part of various fields, including calculus, algebra, and statistics. One of the most common optimization problems is finding the maximum or minimum value of a function. In this article, we will explore the problem of finding the maximum value of the product $P = xy$ when $y = 5 - 2x$. This problem is a classic example of a constrained optimization problem, where we need to find the maximum value of a function subject to a given constraint.

Understanding the Problem

The problem states that we need to find the maximum value of the product $P = xy$ when $y = 5 - 2x$. This means that we are given a function $y = 5 - 2x$, and we need to find the maximum value of the product $P = xy$ subject to this constraint. In other words, we need to find the value of $x$ that maximizes the product $P = xy$.

The Product Function

The product function $P = xy$ is a simple function that represents the product of two variables $x$ and $y$. To find the maximum value of this function, we need to substitute the given constraint $y = 5 - 2x$ into the product function. This will give us a new function in terms of $x$ alone, which we can then optimize.

Substituting the Constraint

Substituting the constraint $y = 5 - 2x$ into the product function $P = xy$, we get:

$$P = x(5 - 2x)$$

Expanding this expression, we get:

$$P = 5x - 2x^2$$

This is a quadratic function in terms of $x$, which we can optimize using calculus.

Finding the Maximum Value

To find the maximum value of the function $P = 5x - 2x^2$, we need to find the critical points of the function. Critical points occur when the derivative of the function is equal to zero or undefined. In this case, the derivative of the function is:

$$\frac{dP}{dx} = 5 - 4x$$

Setting this derivative equal to zero, we get:

$$5 - 4x = 0$$

Solving for $x$, we get:

$$x = \frac{5}{4}$$

This is the critical point of the function. To determine whether this point corresponds to a maximum or minimum, we need to examine the second derivative of the function.

The Second Derivative

The second derivative of the function $P = 5x - 2x^2$ is:

$$\frac{d^2P}{dx^2} = -4$$

Since the second derivative is negative, we know that the critical point $x = \frac{5}{4}$ corresponds to a maximum value of the function.

Finding the Maximum Value

To find the maximum value of the function, we need to substitute the critical point $x = \frac{5}{4}$ into the function $P = 5x - 2x^2$. This gives us:

$$P = 5\left(\frac{5}{4}\right) - 2\left(\frac{5}{4}\right)^2$$

Simplifying this expression, we get:

$$P = \frac{25}{4} - \frac{25}{8}$$

$$P = \frac{25}{8}$$

This is the maximum value of the product $P = xy$ when $y = 5 - 2x$.

Conclusion

In this article, we explored the problem of finding the maximum value of the product $P = xy$ when $y = 5 - 2x$. We used calculus to find the critical points of the function and determined that the critical point $x = \frac{5}{4}$ corresponds to a maximum value of the function. We then found the maximum value of the function by substituting the critical point into the function. The maximum value of the product $P = xy$ when $y = 5 - 2x$ is $\frac{25}{8}$.

Applications

This problem has many applications in various fields, including economics, engineering, and physics. For example, in economics, the product function $P = xy$ can represent the total revenue of a company, where $x$ is the number of units sold and $y$ is the price per unit. In this case, the maximum value of the product function represents the maximum revenue that the company can achieve.

Final Thoughts

In conclusion, finding the maximum value of the product $P = xy$ when $y = 5 - 2x$ is a classic example of a constrained optimization problem. We used calculus to find the critical points of the function and determined that the critical point $x = \frac{5}{4}$ corresponds to a maximum value of the function. We then found the maximum value of the function by substituting the critical point into the function. The maximum value of the product $P = xy$ when $y = 5 - 2x$ is $\frac{25}{8}$. This problem has many applications in various fields and is an important concept in mathematics and optimization.

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Frequently Asked Questions

In the previous article, we explored the problem of finding the maximum value of the product $P = xy$ when $y = 5 - 2x$. In this article, we will answer some of the most frequently asked questions related to this problem.

Q: What is the product function $P = xy$?

A: The product function $P = xy$ is a simple function that represents the product of two variables $x$ and $y$.

Q: What is the constraint $y = 5 - 2x$?

A: The constraint $y = 5 - 2x$ is a given condition that relates the variables $x$ and $y$. In this case, $y$ is expressed in terms of $x$.

Q: How do we find the maximum value of the product function $P = xy$ when $y = 5 - 2x$?

A: To find the maximum value of the product function $P = xy$ when $y = 5 - 2x$, we need to substitute the constraint $y = 5 - 2x$ into the product function. This will give us a new function in terms of $x$ alone, which we can then optimize.

Q: What is the critical point of the function $P = 5x - 2x^2$?

A: The critical point of the function $P = 5x - 2x^2$ is the value of $x$ that makes the derivative of the function equal to zero. In this case, the critical point is $x = \frac{5}{4}$.

Q: Is the critical point $x = \frac{5}{4}$ a maximum or minimum?

A: The critical point $x = \frac{5}{4}$ is a maximum, since the second derivative of the function is negative.

Q: What is the maximum value of the product $P = xy$ when $y = 5 - 2x$?

A: The maximum value of the product $P = xy$ when $y = 5 - 2x$ is $\frac{25}{8}$.

Q: What are some of the applications of this problem?

A: This problem has many applications in various fields, including economics, engineering, and physics. For example, in economics, the product function $P = xy$ can represent the total revenue of a company, where $x$ is the number of units sold and $y$ is the price per unit.

Q: How can I use calculus to solve optimization problems like this one?

A: To use calculus to solve optimization problems like this one, you need to find the critical points of the function and determine whether they correspond to a maximum or minimum. You can do this by examining the first and second derivatives of the function.

Q: What are some of the key concepts in calculus that I need to know to solve optimization problems like this one?

A: Some of the key concepts in calculus that you need to know to solve optimization problems like this one include the derivative, the second derivative, and critical points.

Additional Resources

If you want to learn more about optimization problems like this one, here are some additional resources that you may find helpful:

• Calculus textbooks: There are many calculus textbooks that cover optimization problems in detail. Some popular ones include "Calculus" by Michael Spivak and "Calculus: Early Transcendentals" by James Stewart.
• Online resources: There are many online resources that provide tutorials and examples of optimization problems, including Khan Academy and MIT OpenCourseWare.
• Practice problems: Practice problems are a great way to learn and improve your skills in solving optimization problems. You can find practice problems in calculus textbooks or online resources.

Conclusion

In this article, we answered some of the most frequently asked questions related to the problem of finding the maximum value of the product $P = xy$ when $y = 5 - 2x$. We hope that this article has been helpful in providing you with a better understanding of this problem and how to solve it using calculus. If you have any further questions or need additional help, please don't hesitate to ask.