Problem 25 - Algebra and Trigonometry with Analytic Geometry

Question

A farmer plans to use 180 feet of fencing to enclose a rectangular region, using part of a straight river bank instead of fencing as one side of the rectangle, as shown in the figure on the next page. Find the area of the region if the length of the side parallel to the river bank is (a) twice the length of an adjacent side. (b) one-half the length of an adjacent side. (c) the same as the length of an adjacent side.

Step-by-Step Solution

Answer

(a) 4050 sq ft, (b) 2592 sq ft, (c) 3600 sq ft.

Step 1: Understanding the Problem

The problem involves using 180 feet of fencing to enclose a rectangular area with one side being the river bank. The river bank serves as one side which needs no fencing. We will use the remaining fencing to enclose the other three sides.

Step 2: Formulating Equations for (a)

For case (a), let the length of the side parallel to the river bank be denoted as \( l = 2w \), where \( w \) is the width. The total fencing available is 180 feet, so we use it on the two widths and one length: \( 2w + 2w = 180 \). Simplifying, we get \( 4w = 180 \).

Step 3: Solving for Width in (a)

In the equation \( 4w = 180 \), solve for \( w \) by dividing both sides by 4: \( w = 45 \) feet.

Step 4: Calculating Length in (a)

Since \( l = 2w \), and \( w = 45 \), we find \( l = 2 \times 45 = 90 \) feet.

Step 5: Finding Area for (a)

The area \( A \) of the rectangle is \( l \times w \). Substituting the values \( l = 90 \) and \( w = 45 \), we get \( A = 90 \times 45 = 4050 \) square feet.

Step 6: Formulating Equations for (b)

For case (b), let \( l = \frac{1}{2}w \). Using the same fencing approach, \( 2w + \frac{1}{2}w = 180 \). Simplify to \( 2.5w = 180 \).

Step 7: Solving for Width in (b)

Solving \( 2.5w = 180 \), we find \( w = \frac{180}{2.5} = 72 \) feet.

Step 8: Calculating Length in (b)

\( l = \frac{1}{2}w \), thus \( l = \frac{1}{2} \times 72 = 36 \) feet.

Step 9: Finding Area for (b)

The area \( A = l \times w = 36 \times 72 = 2592 \) square feet.

Step 10: Formulating Equations for (c)

For case (c), \( l = w \). Using the fencing equation: \( l + 2w = 180 \). Replace \( l \) with \( w \), so \( w + 2w = 180 \). Simplifying gives \( 3w = 180 \).

Step 11: Solving for Width in (c)

Solving \( 3w = 180 \), we get \( w = \frac{180}{3} = 60 \) feet.

Step 12: Calculating Length for (c)

Since \( l = w \), \( l = 60 \) feet as well.

Step 13: Finding Area for (c)

The area \( A = l \times w = 60 \times 60 = 3600 \) square feet.

Key Concepts

Perimeter and Area

When it comes to geometry, understanding perimeter and area is essential, especially in real-life applications like the fencing problem. The perimeter is the distance around a shape, and in our problem, it refers to the 180 feet of fencing available to enclose the rectangular area. Since a river bank forms one side of the rectangle, you only need fencing for the remaining three sides. This changes how we calculate the perimeter.

The area is the space within a shape, measured in square units. In our exercise, the area is determined by multiplying the length and width of the rectangle. Knowing both the perimeter to be used and the desired relationship between the length and width helps solve for these dimensions and subsequently calculate the area.

Understanding both concepts allows us to adapt to different scenarios, such as when the length is a multiple of the width, affects the area calculation.

Fencing Problem

The fencing problem is a practical application of geometry and algebra where the goal is to optimize or determine dimensions given a fixed perimeter. In our scenario, with one side of the rectangle being a river, the problem simplifies as you need less fencing. The challenge is figuring out the optimal length and width to maximize or find the required area while only using 180 feet of fencing.

This problem illustrates the use of mathematical reasoning to find the best solution given constraints. Different relationships between the length and width—such as one being twice the other, equal, or half—demonstrate how altering parameters affects the area. Through logic and equations, you can explore these variations to find the correct dimensions.

Algebraic Equations

Algebraic equations are a crucial tool in solving the fencing problem. They help translate a word problem into mathematical expressions that can be solved numerically. For instance, if the length is twice the width, you can express it as the equation: \( l = 2w \).

Using available fencing, you know that two widths and one length must sum up to 180, allowing us to set up an equation like \( 2w + l = 180 \). Simplifying and solving these equations enables you to determine each dimension. Through steps like dividing or multiplying, algebra provides a clear path from problem to solution.

Understanding how to write and manipulate these equations is vital. It turns geometric scenarios into solvable problems, giving clarity and ensuring you don't exceed the assumptions like the total fencing available.

Geometry in Real Life

Geometry is not just a theoretical subject; it finds its place in solving real-life problems, such as determining land use in agriculture. In the fencing problem, you must use your knowledge of shapes, space, and formulas to find solutions that meet practical needs.

Farmers and property managers often encounter such challenges, needing to partition land efficiently. Finding the maximum area they can enclose with a set amount of resources, like fencing, helps optimize land usage. This is where the abstract concepts of geometry and the precision of algebra intersect to support everyday decision-making.

These applications show the relevance and importance of math in everyday life, beyond the classroom. Understanding how to apply these concepts effectively can lead to better resource management and planning.