Maximizing Area with a Fixed Perimeter: A Comprehensive Guide

When tasked with using a fixed amount of fencing material to enclose a rectangular plot of land, determining the optimal dimensions to maximize the enclosed area is a fundamental problem in geometry and optimization. This article delves into the process of finding the ideal length and width for different scenarios, focusing on how to maximize area with a given perimeter.

Introduction to Perimeter and Area Optimization

In this article, we will explore the mathematical concepts and methods used to determine the dimensions of a rectangular plot that maximizes the enclosed area, given a fixed perimeter. We will walk through a detailed process, providing clear steps and examples to help you understand the optimization process.

Step-by-Step Optimization: A Detailed Process

Step 1: Define the Variables

Let (l) length and (w) width of the rectangle.

Step 2: Set Up the Perimeter Equation

The perimeter (P) of a rectangle is given by:

[P 2l 2w]

Given that the total perimeter is 448 feet:

[2l 2w 448]

Step 3: Simplify the Perimeter Equation

Dividing the entire equation by 2:

[l w 224]

From this, we can express (w) in terms of (l):

[w 224 - l]

Step 4: Set Up the Area Equation

The area (A) of a rectangle is given by:

[A l times w]

Substituting the expression for (w):

[A l (224 - l) 224l - l^2]

Step 5: Maximize the Area

To find the maximum area, we take the derivative of (A) with respect to (l) and set it to zero:

[frac{dA}{dl} 224 - 2l]

Setting the derivative equal to zero:

[224 - 2l 0]

Solving for (l):

[2l 224 implies l 112]

Step 6: Find the Width

Using (l 112) to find (w):

[w 224 - l 224 - 112 112]

Conclusion:

The dimensions that maximize the area of the rectangular plot are:

- Length (l 112) feet - Width (w 112) feet

Thus, the optimal shape is a square with dimensions of (112) feet by (112) feet. The maximum area enclosed will be:

[A 112 times 112 12544 text{ square feet}]

Additional Examples and Scenarios

Example 1: A Smaller Enclosure

Consider a smaller rectangular plot with a fixed perimeter of 200 feet. If the dimensions are 50 feet by 50 feet, the enclosed area is 2500 square feet.

Example 2: Optimization with a Given Area

Let (X) be the length of the rectangle and (Y) be the width. Given the area of the enclosure is 4100 square feet and the perimeter is 450 feet:

(XY 4100)

(2X 2Y 450)

From the area equation, we have:

(Y frac{4100}{X})

Substituting into the perimeter equation:

(X frac{4100}{X} 225)

Multiplying through by (X) results in a quadratic equation:

(X^2 - 225X 4100 0)

Solving the quadratic equation:

(X 20) or (X 205)

Thus, the dimensions are either 20 feet by 205 feet or 205 feet by 20 feet, and the area is:

(20 times 205 4100 text{ square feet})

Conclusion and Advice

For the optimal rectangle that maximizes the enclosed area with a fixed perimeter, a square is generally the best shape, as it provides the largest area for a given perimeter. In our example, the dimensions that maximize the area are 112 feet by 112 feet, resulting in a maximum area of 12,544 square feet.

Frequently Asked Questions

Q: What is the optimal shape for maximizing the area with a fixed perimeter?

A: The optimal shape is a square. A square provides the largest area for a given perimeter.

Q: How do you find the dimensions of a rectangle with a given perimeter to maximize the area?

A: By using calculus, specifically by taking the derivative of the area function with respect to length and setting it to zero. Alternatively, using the relationship between area and perimeter, as shown in the examples above.

Q: Can the length and width of a rectangle be different to maximize the area?

A: Yes, but for a fixed perimeter, a square will provide the maximum area. If the dimensions can vary, different lengths and widths can maximize the area depending on the specific criteria.