Problem 14 - Beginning and Intermediate Algebra

Question

Solve. Find the dimensions of the rectangular garden of greatest area that can be enclosed with \(40 \mathrm{ft}\) of fencing.

Step-by-Step Solution

Answer

The dimensions of the rectangular garden of greatest area that can be enclosed with 40ft of fencing are 10ft by 10ft, making it a square.

Step 1: Express one dimension in terms of the other

Given that 2L + 2W = 40, let's express L in terms of W (or vice versa): \(L = 20 - W\) Step 2: Define the area function in terms of one variable

Step 2: Express the area in terms of W

Replace L in the area function with its equivalent expression from Step 1: \(A(W) = W(20 - W)\) Step 3: Find the critical points

Step 3: Differentiate A(W) and set it equal to 0

Find the first derivative of the area function with respect to W: \(A'(W) = 20 - 2W\) Now, set A'(W) equal to 0 to find the critical points: \(20 - 2W = 0\) Step 4: Solve for W

Step 4: Find the value of W

Solve the equation from Step 3 for W: \(W = 10\) Step 5: Calculate the value of L

Step 5: Calculate the length

Use the value of W and our initial expression for L to determine the length: \(L = 20 - W = 20 - 10 = 10\) Step 6: Verify and report the result

Step 6: Verify the result and write the solution

Since both L = 10 and W = 10, the garden of greatest area that can be enclosed with 40ft of fencing is a square with dimensions 10ft by 10ft.

Key Concepts

Rectangular Garden

When designing a garden, especially a rectangular one, understanding the layout is essential. A rectangular garden has two key dimensions, the length (L) and the width (W). These parameters define the shape and ultimately the area you can plant within. In our exercise, the challenge is to find dimensions that maximize the area of the garden using a fixed perimeter of fencing. Whether for amateur gardeners or seasoned landscapers, figuring out these dimensions holds practical value. A well-planned rectangular garden not only optimizes usable space but also makes efficient use of resources like fencing material.

Perimeter and Area

Knowing how to calculate the perimeter and area of a rectangular garden is vital for optimizing its layout. The perimeter is the total length of fencing needed to enclose a garden and is calculated by the formula:

• Perimeter = 2L + 2W

In this problem, the perimeter is fixed at 40 feet.
To solve for one dimension in terms of the other, you can rearrange the formula, such as expressing L as a function of W, which leads to:

• \(L = 20 - W\)

Then, the area of the garden is defined by:

• Area = L × W

Substitute the expression for L obtained from the perimeter into the area formula to express the area purely in terms of one variable, W:

• \(A(W) = W(20 - W)\)

This quadratic equation represents how the area changes as W changes, helping us find the dimensions that maximize this area.

Critical Points

Critical points are key in optimization problems as they often reveal the maximum or minimum values a function can take. In our context, critical points help find the maximum area of the garden. To identify these points, you first need the first derivative of the area function, as it shows how the area changes with respect to width, W.
The critical points occur where this derivative equals zero:

• \(A'(W) = 0\)
• \(20 - 2W = 0\)

by solving, we find

• W = 10

This means that when the width is 10 feet, one potential critical point occurs. These critical points help in confirming where the area is maximized, which is essential to find the dimensions that use the fencing most efficiently.

First Derivative Test

The first derivative test is a powerful tool to determine if the critical points found are maximum, minimum, or points of inflection. Once you have a critical point from the area function, like W = 10, you need to apply this test. By evaluating the sign of the derivative before and after the critical point, you can infer the nature of this point.
If the derivative changes from positive to negative as it passes through the critical point, the function is increasing before and decreasing after, confirming a local maximum, which for our context is the garden's greatest area.
Since

• \(A'(W) = 20 - 2W\)

safeguards that as W increases from just below 10 to just above 10, the derivative switches from positive to negative, indicating a maximum area value at this point.
This ensures that the dimensions of the rectangular garden, 10 feet by 10 feet, yield the largest possible area within the constraint of a 40-foot perimeter.