Farmer John has 20 meters of fencing and wants to enclose the largest possible rectangular field. This problem may seem simple, but understanding the underlying mathematical principles can lead to an efficient, practical solution. In this article, we will explore the process of maximizing the area of a rectangle with a fixed perimeter using basic algebra and geometry concepts. Let's dive into the solution step-by-step.

Understanding the Problem

A rectangle's perimeter can be expressed using the formula:

P 2L 2W

where L is the length and W is the width. Given that the fence has a fixed length of 20 meters, we can set up the equation as follows:

2L 2W 20

Expressing the Area

The area A of a rectangle is given by:

A L x W

To maximize the area, we need to express one variable in terms of the other. We can rearrange the perimeter equation to find W in terms of L as follows:

2W 20 - 2L

Dividing both sides by 2, we get:

W 10 - L

Now we substitute W 10 - L into the area formula:

A L x (10 - L)

Simplifying, we obtain:

A 10L - L^2

Finding the Optimal Dimensions

The equation A 10L - L^2 represents a quadratic function, which forms a parabola when plotted. Since the coefficient of L^2 is negative, the parabola opens downwards, and its maximum point (vertex) gives the maximum area. The vertex of a parabola defined by A ax^2 bx c can be found using the formula:

x -b / 2a

Here, a -1 and b 10. Therefore, the length L that maximizes the area is:

L -10 / (2 * -1) 5 meters

Calculating the Width and Area

Substituting L 5 meters into the equation W 10 - L, we get:

W 10 - 5 5 meters

Thus, the dimensions of the field are:

Length (L) 5 meters

Width (W) 5 meters

The maximum area is then:

A 5 meters x 5 meters 25 square meters

Conclusion

Farmer John can create the largest rectangular field with 20 meters of fencing by making the field a square with each side measuring 5 meters. This solution can be generalized to any similar problem where the goal is to maximize the area of a rectangle with a fixed perimeter. By understanding the relationship between length and width, we can optimize the use of limited resources to achieve the desired outcome.

Key Takeaways

To maximize the area of a rectangle with a fixed perimeter, the length and width should be equal for a square shape. The optimal dimensions for the rectangular field are 5 meters by 5 meters, yielding a maximum area of 25 square meters. This method can be applied to similar problems where the goal is to optimize area with limited resources.

By leveraging basic algebra and geometry, we’ve solved a practical problem in agriculture while providing a simple yet effective approach to resource optimization.