Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. A farmer has 1000 feet of fence to enclose a rectangular area. What dimensions for the rectangle result in the maximum area enclosed by the fence?

Step-by-Step Solution

Answer

The rectangle with maximum area is a square with sides of 250 feet.

Step 1: Understand the problem

We need to maximize the area of a rectangle with a fixed perimeter of 1000 feet. A perimeter P of a rectangle is given by the formula \(P = 2L + 2W\), where \(L\) is the length and \(W\) is the width.

Step 2: Express width in terms of length

Given the perimeter equation \(2L + 2W = 1000\), solve for \(W\). Simplifying gives:\[W = 500 - L\]

Step 3: Write the area function

The area \(A\) of a rectangle is given by \(A = L \times W\). Substitute \(W = 500 - L\) into the area formula:\[A = L(500 - L) = 500L - L^2\]

Step 4: Find the critical point

To maximize the area, take the derivative of the area function \(A = 500L - L^2\) with respect to \(L\) and set it to zero:\[\frac{dA}{dL} = 500 - 2L = 0\]Solve for \(L\): \[L = 250\]

Step 5: Calculate the width

Substitute \(L = 250\) back into the equation \(W = 500 - L\):\[W = 500 - 250 = 250\]

Step 6: Conclusion

The dimensions of the rectangle that maximize the area are 250 feet by 250 feet, which is a square.

Key Concepts

Rectangular Perimeter Problem

The Rectangular Perimeter Problem is a classic optimization problem in calculus that involves finding the best way to use a limited amount of resources—in this case, fencing—to enclose the largest possible area. When given a fixed perimeter to work with, the challenge is to determine the length and width of a rectangle that will maximize the enclosed area.
The perimeter of a rectangle is calculated using the formula:

• Perimeter formula: \( P = 2L + 2W \)
• Where \( L \) is the length and \( W \) is the width.

In our problem, the farmer has a total of 1000 feet of fence available. This means that whatever dimensions we choose for our rectangle, they must satisfy the equation \( 2L + 2W = 1000 \). By simplifying this equation, we can express one variable in terms of the other, which is often a helpful first step in solving optimization problems.
All that remains now is to maximize the area using the available perimeter efficiently.

Area Maximization

Area Maximization focuses on using calculus techniques to find the largest possible area that can be enclosed given the constraints of a problem. For rectangles, the area \( A \) is calculated as:

• Area formula: \( A = L \times W \)

After expressing the width in terms of the length, \( W = 500 - L \), the area function becomes \( A = L(500 - L) \), which simplifies to \( A = 500L - L^2 \). This quadratic equation describes how area changes with respect to the rectangle's length, \( L \).
The goal here is to find the value of \( L \) that will produce the maximum area. This is achieved by identifying the critical points—values of \( L \) where the slope of the area function is zero—using calculus.

Critical Points in Calculus

Critical Points in Calculus are essential for solving optimization problems. They are points at which the derivative of a function equals zero, or where the derivative does not exist. These points often correspond to maximum or minimum values of a function.
Within the context of the rectangular area problem, we need to find the critical points of the area function \( A = 500L - L^2 \).
To find these critical points, we take the derivative of the area function with respect to \( L \) and set it to zero:

• Derivative of area: \( \frac{dA}{dL} = 500 - 2L \)
• Set the derivative to zero: \( 500 - 2L = 0 \)
• Solve for \( L \): \( L = 250 \)

Thus, the critical point occurs when \( L = 250 \). When we substitute this back, we find \( W = 250 \), leading to a square. This is the configuration that maximizes the enclosed area, illustrating how calculus helps us solve such real-world optimization challenges.