Rectangle Area: Max Perimeter (8th Grade)

Hey guys! Let's dive into a fun math problem that involves rectangles, areas, perimeters, and a bit of thinking. This is a classic problem often seen in 8th-grade math, especially when you're learning about factors and multiples. So, let's break it down step-by-step.

Understanding the Problem

Our problem states: A rectangle has an area of 25 square meters. What is the maximum possible perimeter of this rectangle, and what will be the result in centimeters?

To solve this, we need to understand the relationship between the area and perimeter of a rectangle. Remember:

• Area of a rectangle = length × width
• Perimeter of a rectangle = 2 × (length + width)

The key here is that the area is fixed (25 square meters), but the length and width can vary. We need to find the length and width that give us the maximum perimeter.

Finding Possible Length and Width Combinations

Since the area is 25 square meters, we need to find pairs of numbers that multiply to give 25. Let's list them out:

1. Length = 1 meter, Width = 25 meters (1 × 25 = 25)
2. Length = 5 meters, Width = 5 meters (5 × 5 = 25)

These are the most straightforward combinations using whole numbers. Of course, we could have decimal or fractional lengths and widths that also multiply to 25 (like 2.5 and 10), but let's stick to these simpler options for now to understand the concept. Remember, focusing on integer values for length and width is crucial, especially in the context of 8th-grade math problems dealing with factors and multiples. By considering integer pairs, we can easily identify the combinations that yield the maximum perimeter.

Why These Combinations?

It's essential to grasp why we're focusing on pairs that multiply to 25. The area of a rectangle is fundamentally the product of its length and width. By identifying these pairs, we're essentially exploring all possible rectangular shapes that can enclose an area of exactly 25 square meters. Each pair represents a unique configuration, and our goal is to determine which configuration results in the largest possible perimeter. This involves understanding how changes in length and width affect the overall perimeter while maintaining a constant area. Furthermore, this exercise reinforces the concept of factors and multiples, as we're essentially finding factors of 25 and using them to construct rectangles with the specified area. This approach provides a solid foundation for understanding the relationship between area, perimeter, and the dimensions of a rectangle, which is a fundamental concept in geometry.

Calculating the Perimeters

Now, let's calculate the perimeter for each of these combinations:

1. Length = 1 meter, Width = 25 meters: Perimeter = 2 × (1 + 25) = 2 × 26 = 52 meters
2. Length = 5 meters, Width = 5 meters: Perimeter = 2 × (5 + 5) = 2 × 10 = 20 meters

From these calculations, we can see that the combination of length = 1 meter and width = 25 meters gives us the maximum perimeter of 52 meters.

Maximizing Perimeter with Fixed Area

Notice something interesting: when the length and width are very different (like 1 and 25), the perimeter is larger. When the length and width are closer together (like 5 and 5 – a square!), the perimeter is smaller. This is a general principle: for a fixed area, the perimeter is maximized when the rectangle is long and thin. This concept is important for understanding optimization problems in mathematics. Thinking about different shapes with the same area can help visualize why a more elongated shape has a larger perimeter. Understanding this relationship can also aid in solving practical problems involving minimizing materials used to enclose a certain area.

Converting to Centimeters

The problem asks for the answer in centimeters. We know that 1 meter = 100 centimeters. So, to convert 52 meters to centimeters, we multiply by 100:

52 meters × 100 centimeters/meter = 5200 centimeters

Therefore, the maximum possible perimeter of the rectangle is 5200 centimeters.

The Answer

The maximum possible perimeter of a rectangle with an area of 25 square meters is 5200 centimeters.

Key Takeaways

• Area and Perimeter Relationship: Understanding how the area and perimeter of a rectangle are related is crucial.
• Factors and Multiples: This problem reinforces the concept of factors and multiples.
• Maximization: For a fixed area, the perimeter is maximized when the length and width are as different as possible.
• Unit Conversion: Don't forget to convert units when necessary!

Real-World Application

This type of problem isn't just theoretical. Imagine you're a farmer who wants to enclose a rectangular field with 25 square meters of area using the least amount of fencing. Knowing how to minimize the perimeter for a given area helps you save money on fencing materials!

Exploring Further: Non-Integer Values

As mentioned earlier, we initially focused on integer values for the length and width. What happens if we consider non-integer values? Let's explore a few examples to see how the perimeter changes.

Example 1: Length = 2.5 meters, Width = 10 meters

• Area = 2.5 meters × 10 meters = 25 square meters
• Perimeter = 2 × (2.5 meters + 10 meters) = 2 × 12.5 meters = 25 meters

Converting to centimeters: 25 meters × 100 centimeters/meter = 2500 centimeters

Example 2: Length = 0.5 meters, Width = 50 meters

• Area = 0.5 meters × 50 meters = 25 square meters
• Perimeter = 2 × (0.5 meters + 50 meters) = 2 × 50.5 meters = 101 meters

Converting to centimeters: 101 meters × 100 centimeters/meter = 10100 centimeters

As you can see, by using non-integer values, particularly when the length and width are significantly different, we can achieve even larger perimeters. This illustrates that the maximum perimeter is achieved as one side approaches zero and the other approaches infinity (while maintaining the constant area). This exploration demonstrates that the initial restriction to integer values was for simplicity in understanding the core concept, but the principle extends to all real numbers.

Implications for Calculus

For those who are familiar with calculus, this problem can be approached using optimization techniques. By expressing the perimeter as a function of one variable (either length or width) and using derivatives, you can find the critical points and determine the dimensions that maximize the perimeter. This provides a more rigorous mathematical approach to the problem and reinforces the understanding of optimization principles.

Conclusion

So there you have it! We've successfully found the maximum possible perimeter of a rectangle with a given area. Remember the key concepts and you'll be able to tackle similar problems with confidence. Keep practicing, and you'll become a math whiz in no time!