Find Two Positive Real Numbers Whose Product Is A Maximum

Finding Two Positive Real Numbers Whose Product is a Maximum: A Comprehensive Guide

Finding the maximum product of two positive real numbers is a classic optimization problem with applications across various fields, from geometry and physics to economics and computer science. This seemingly simple problem provides a great introduction to the power of calculus and its applications in finding optimal solutions. This comprehensive guide will delve into the problem, exploring different approaches and providing a thorough understanding of the underlying concepts.

Understanding the Problem

The core of the problem is stated as follows: Given a sum of two positive real numbers, find the two numbers such that their product is maximized. Let's represent these two numbers as 'x' and 'y'. We are given that x + y = S, where S is a constant representing the sum of the two numbers. Our goal is to maximize the product P = x * y.

Method 1: Using Calculus

This is the most common and elegant method for solving this problem. We'll use calculus to find the critical points of the function representing the product and then determine whether these points correspond to a maximum.

1. Expressing 'y' in terms of 'x'

Since x + y = S, we can express 'y' in terms of 'x': y = S - x.

2. Defining the Product Function

Now we can rewrite the product function P in terms of 'x' only:

P(x) = x * (S - x) = Sx - x²

3. Finding the Derivative

To find the critical points, we take the derivative of P(x) with respect to 'x':

P'(x) = S - 2x

4. Setting the Derivative to Zero

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

S - 2x = 0 2x = S x = S/2

5. Finding the Second Derivative

To confirm that this critical point represents a maximum, we examine the second derivative:

P''(x) = -2

Since the second derivative is negative, the critical point corresponds to a maximum.

6. Finding 'y'

Substituting x = S/2 back into the equation y = S - x, we get:

y = S - (S/2) = S/2

7. Conclusion

Therefore, the maximum product is achieved when x = S/2 and y = S/2. The maximum product itself is (S/2) * (S/2) = S²/4.

Method 2: Using the AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) inequality provides a powerful and elegant alternative solution. The AM-GM inequality states that for non-negative real numbers a and b:

(a + b) / 2 ≥ √(ab)

Equality holds if and only if a = b.

1. Applying the AM-GM Inequality

In our case, a = x and b = y. We have:

(x + y) / 2 ≥ √(xy)

Since x + y = S, we can rewrite this as:

S / 2 ≥ √(xy)

2. Squaring Both Sides

Squaring both sides, we get:

S²/4 ≥ xy

3. Maximum Product

This inequality shows that the product xy is always less than or equal to S²/4. Equality holds when x = y, which implies x = y = S/2.

4. Conclusion

Therefore, the maximum product is achieved when x = S/2 and y = S/2, resulting in a maximum product of S²/4. This method offers a concise and insightful approach to the problem.

Geometric Interpretation

The problem can also be visualized geometrically. Consider a rectangle with sides x and y. The perimeter of the rectangle is 2(x + y) = 2S, which is constant. The area of the rectangle represents the product xy. To maximize the area (product), we need to form a square where x = y = S/2. This square has the largest area compared to any other rectangle with the same perimeter.

Applications

This seemingly simple optimization problem has wide-ranging applications:

• Geometry: Finding the dimensions of a rectangle with a fixed perimeter that maximizes its area.
• Physics: Optimizing the dimensions of physical structures to maximize certain properties.
• Economics: Determining optimal allocation of resources to maximize profits.
• Computer Science: Algorithm design and optimization problems.

Extending the Problem

The problem can be extended to consider more than two numbers. For example, finding three positive real numbers whose sum is S and whose product is maximized. The solution involves similar techniques, leading to the conclusion that the maximum product occurs when all three numbers are equal to S/3. This generalizes to n numbers, where the maximum product occurs when all numbers are equal to S/n.

Frequently Asked Questions (FAQs)

Q: What if the numbers are not necessarily positive?

A: If the numbers can be negative, the problem becomes more complex and doesn't necessarily have a maximum product. The product can become arbitrarily large in the negative direction.

Q: Can this problem be solved without calculus?

A: Yes, the AM-GM inequality provides a calculus-free solution. Completing the square on the quadratic function P(x) = Sx - x² also provides a non-calculus solution.

Q: What are some real-world examples of this problem?

A: Imagine a farmer with a fixed amount of fencing wanting to enclose the largest possible rectangular area. Or a manufacturer wanting to minimize packaging material while maximizing the volume of the product inside. These scenarios mirror the core concept of this problem.

Conclusion

Finding two positive real numbers whose product is a maximum is a fundamental optimization problem with a straightforward solution using calculus or the AM-GM inequality. The problem's elegant solution and its diverse applications highlight the power of mathematical tools in solving real-world problems across various fields. Understanding this problem provides a solid foundation for tackling more complex optimization challenges in the future. The geometric interpretation provides an intuitive understanding, reinforcing the mathematical results and showcasing the interconnectedness of different mathematical concepts. The problem's simplicity belies its significant importance in various applications, making it a valuable concept to grasp for anyone interested in mathematics, science, or engineering.