Maximizing the Area of a Rectangle with a Given Perimeter

Introduction to Perimeter and Area:

Understanding the relationship between the perimeter and area of a rectangle is crucial in various mathematical and practical applications. In this article, we will explore how to find the maximum area of a rectangle when its perimeter is fixed at 72 cm. This exploration will involve the application of basic algebra and the principles of optimization.

The Perimeter and Its Impact on Area

The perimeter of a rectangle is the total length of its boundaries. Given that a perimeter of 72 cm can be expressed as P 72 cm, we need to determine the dimensions that will yield the maximum area. Let's denote the length of the rectangle as l and the width as w. The perimeter is given by:

P 2l 2w

Given the fixed perimeter, we can simplify this to:

2l 2w 72

Dividing by 2, we get:

l w 36

Now, we need to find the area of the rectangle, which is given by:

A l times w

Optimization of the Area

To maximize the area, we express one variable in terms of the other. From the perimeter equation:

w 36 - l

Substituting this into the area formula, we get:

A l times (36 - l) 36l - l^2

This is a quadratic equation in the standard form:

A -l^2 36l

For such a quadratic equation, the maximum value occurs at the vertex. The vertex form can be found using the formula for the x-coordinate, where the maximum area is:

l -frac{b}{2a}

Here, a -1 and b 36. Plugging in the values:

l -frac{36}{2 times -1} 18 text{ cm}

Using this value, we can find the width:

w 36 - 18 18 text{ cm}

Therefore, the dimensions of the rectangle with maximum area are 18 cm by 18 cm, which makes it a square. The maximum area is:

A 18 times 18 324 text{ cm}^2

Generalization of the Concept

The concept of maximizing a rectangle's area with a fixed perimeter can be generalized to other perimeters as well. Using the same method:

text{If the perimeter is} ; 72 text{ cm, then} ; l w 36

Expressing w in terms of e:

w 36 - l implies A l times (36 - l) 36l - l^2

Thus, the area is maximized when l 18 text{ cm and} ; w 18 text{ cm}, making the rectangle a square and the maximum area 324 text{ cm}^2.

Conclusion

In summary, to maximize the area of a rectangle with a fixed perimeter, the rectangle must be a square. For a perimeter of 72 cm, the maximum area achievable is 324 cm2. This principle is useful in various fields, including architecture, engineering, and everyday design problems.