Question

THOUGHT PROVOKING You have 100 feet of fencing to enclose a rectangular garden. Draw three possible designs for the garden. Of these, which has the greatest area? Make a conjecture about the dimensions of the rectangular garden with the greatest possible area. Explain your reasoning.

Short answer

The square with length and width of 25 feet respectively has the greatest area of 625 sq. feet.

Step-by-step solution

Understanding the situation

The problem suggests that there is a fencing of 100 feet that needs to create a rectangular garden. Therefore, the perimeter of the rectangle, which is two times the sum of the length and the width, is 100 feet.

Define Possible Rectangles

To find the rectangles, you can try different values for the length \(L\) and width \(W\), as long as \(2(L + W) = 100\) or \(L + W = 50\). Let's take three examples: a) when both length and width are equal (which makes a square), i.e., \(L = W = 25\) feet; b) when \(L = 30\) feet, and \(W = 20\) feet; c) when \(L = 40\) feet, and \(W = 10\) feet.

Calculate The Area of Each Rectangle

The area of a rectangle is calculated by multiplying length and width. For each rectangle: a) The area is \(25*25 = 625\) sq. feet. b) The area is \(30*20 = 600\) sq. feet. c) The area is \(40*10 = 400\) sq. feet.

Compare Areas

By comparing the areas of these three rectangles, it is clear that the square (rectange a) has the greatest area.

Problem interpretation

One can conjecture that the rectangle with the greatest possible area that could be enclosed with a 100 feet fence, will be a square. This is because the square is the rectangle that has the maximum area for a given perimeter, which is a property of the square.

Key concepts

Perimeter and Area Relationship

When dealing with shapes like rectangles, understanding the relationship between perimeter and area is crucial. The perimeter is the total length around the shape, and for a rectangle, it is calculated by adding up all its sides. The formula for the perimeter of a rectangle is:

• Perimeter = 2(Length + Width)

In this fencing problem, the total available perimeter is 100 feet. Hence, we set up the equation:

• 2(L + W) = 100
• or simplified, L + W = 50

The area of a rectangle, on the other hand, is the amount of space inside the rectangle. We calculate it by multiplying the length by the width:

• Area = Length × Width

Understanding how these two measurements interact helps us explore how changing the dimensions of a rectangle affects its area, given a fixed perimeter.

Rectangle Dimensions

When exploring rectangle dimensions with a fixed perimeter, we need to consider various combinations of length and width that satisfy the perimeter equation. For example, with a total perimeter of 100 feet, the equation becomes:

• L + W = 50

You can select multiple pairs of dimensions:

• A square, where Length ( L ) = Width ( W ) = 25 feet
• A more elongated rectangle, such as Length = 30 feet and Width = 20 feet
• Another possibility, Length = 40 feet and Width = 10 feet

Each of these combinations makes a rectangle but impacts the enclosed area differently. Observing these variations is key in optimizing the area under a fixed perimeter condition. By comparing these different shapes, one can determine which dimensions yield the maximum area.

Fencing Problem

The fencing problem is a classical exercise in optimization where you seek to maximize the area enclosed by a fixed length of fencing, forming a rectangular shape. Given 100 feet of fencing, different configurations of length and width impact the total area.

• Elongated rectangles tend to have smaller areas.
• When the dimensions become equal, forming a square, the area is maximized. For the given length of fencing, when both sides are 25 feet, the area peaks at 625 square feet.

The central conclusion of this problem is a mathematical principle: For a fixed perimeter, a square encloses the largest possible area among all rectangles. This insight underlines why squares are often used in design and layout planning when space optimization is required. By understanding the nature of this problem, students can apply similar reasoning to other geometric and real-world challenges.