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Chapter 3: Problem 100

Cross Fenced Kim wants to construct rectangular pens for four animals with 400 feet of fencing. To get four separate pens she will fence a large rectangle and then fence through the middle of the rectangle parallel to the length and parallel to the width. What overall dimensions will maximize the total area of the pens?

Short Answer

Expert verified
The dimensions to maximize the area are 100 feet by approximately 66.67 feet.

Step by step solution

01

- Define variables

Let the length of the large rectangle be denoted by \( L \) and the width by \( W \). To create four pens, additional fencing is used to divide the large rectangle. Specifically, two sections of the fencing will be parallel to the length (\( L \)), and one section will be parallel to the width (\( W \)).
02

- Express total fencing used

The total fencing required will include the perimeter of the large rectangle plus the additional fencing inside. The total fencing used can be written as: \[ 2L + 3W = 400 \]
03

- Solve for one variable

Solve the fencing equation for one of the variables. Let鈥檚 isolate \( L \): \[ 2L = 400 - 3W \] \[ L = 200 - \frac{3}{2}W \]
04

- Express the area

The area \( A \) of the large rectangle is: \[ A = L \times W \] Substitute \( L \) from the previous step: \[ A = (200 - \frac{3}{2}W) \times W \] \[ A = 200W - \frac{3}{2}W^2 \]
05

- Find the critical points

To maximize the area, take the derivative of \( A \) with respect to \( W \) and set it to zero: \[ \frac{dA}{dW} = 200 - 3W = 0 \] Solving for \( W \): \[ 3W = 200 \] \[ W = \frac{200}{3} \approx 66.67 \]
06

- Find corresponding length

Using the value of \( W \) in the equation for \( L \): \[ L = 200 - \frac{3}{2} \times \frac{200}{3} \] \[ L = 200 - 100 \] \[ L = 100 \]
07

- Verify the dimensions

Verify the total fencing used with these dimensions: \[ 2L + 3W = 2(100) + 3(66.67) \] \[ 2(100) + 3(66.67) = 200 + 200 = 400 \] The total fencing used is indeed 400 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

rectangular pens
When constructing rectangular pens, we need to consider how the shape and the layout of the fencing will affect the total area. For example, if we need to create four separate pens within a larger rectangle, we need additional fencing inside the large rectangle. This extra fencing will be parallel to both the length and width of the rectangle.
In this exercise, we let the length of the large rectangle be denoted by \(L\) and the width by \(W\). To create four pens, additional fencing is used to divide the large rectangle into smaller sections. This extra fencing consists of two sections parallel to the length (\(L\)) and one section parallel to the width (\(W\)).
fencing problem
The fencing problem involves calculating the total length of fencing needed and optimizing the dimensions to achieve the desired outcome. Here, we have a total of 400 feet of fencing for constructing the large rectangle and additional sections within it.
The total fencing needed includes the perimeter of the large rectangle and the additional fencing inside. We can express this relationship as \(2L + 3W = 400\). This equation helps us solve for our required dimensions, considering both external and internal fencing.
area maximization
Area maximization is a key concept in optimization problems. Here, we are looking to maximize the area of the pens using the given amount of fencing. To do this, we first express the area of the large rectangle as \(A = L \times W\).
Next, by substituting \(L\) with the value derived from the fencing equation \(L = 200 - \frac{3}{2}W\), we get \(A = (200 - \frac{3}{2}W) \times W\). Simplifying, we obtain \(A = 200W - \frac{3}{2}W^2\). This new equation represents the area in terms of one variable, \(W\), which will help us find the maximum area.
derivative for optimization
To find the maximum area, we use derivatives. By taking the derivative of \(A\) with respect to \(W\) and setting it to zero, we find the critical points. Differentiating, we get \(\frac{dA}{dW} = 200 - 3W = 0\). Solving for \(W\), we obtain \(W = \frac{200}{3} \text{which is approximately 66.67}\).
Using this value of \(W\), we find the corresponding length by substituting it back into the equation for \(L\): \(L = 200 - \frac{3}{2} \times \frac{200}{3} = 100\).
Finally, we verify the total fencing used: \(2L + 3W = 2(100) + 3(66.67) = 200 + 200 = 400\). This confirms the total fencing used is 400 feet, validating our solution.

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