Problem 7 - Active Calculus

Question

A farmer wants to start raising cows, horses, goats, and sheep, and desires to have a rectangular pasture for the animals to graze in. However, no two different kinds of animals can graze together. In order to minimize the amount of fencing she will need, she has decided to enclose a large rectangular area and then divide it into four equally sized pens by adding three segments of fence inside the large rectangle that are parallel to two existing sides. She has decided to purchase \(7500 \mathrm{ft}\) of fencing. What is the maximum possible area that each of the four pens will enclose?

Step-by-Step Solution

Answer

The maximum possible area that each of the four pens will enclose is 390625 square feet.

Step 1 - Define Variables

Let the length of the large rectangle be denoted by \(L\) and the width by \(W\). Since the large rectangle is divided into four equally sized pens, each pen will have dimensions \(\frac{L}{2} \times \frac{W}{2}\). Let the total amount of fencing available be denoted by \(7500 \mathrm{ft}\).

Step 2 - Write the Equation for the Fencing

The fencing forms the outer rectangle plus three internal segments. Thus, the total amount of fencing used is given by: \[ 2L + 2W + W + L = 3L + 3W \]We know the total fencing is 7500 feet, so: \[ 3L + 3W = 7500 \]

Step 3 - Simplify the Fencing Equation

Divide the equation by 3 to simplify: \[ L + W = 2500 \]

Step 4 - Express Area in Terms of One Variable

The area of the large rectangle is \(L \times W\), and we need to maximize this. Express \(W\) in terms of \(L\) using the equation from Step 3: \[ W = 2500 - L \]The area \(A\) of the rectangle can then be written as: \[ A = L \times W = L \times (2500 - L) = 2500L - L^2 \]

Step 5 - Find the Maximum Area

To find the maximum area, take the derivative of \(A\) with respect to \(L\) and set it to zero: \[ \frac{dA}{dL} = 2500 - 2L \]Set the derivative equal to zero to find the critical points: \[ 2500 - 2L = 0 \]Solving for \(L\) gives: \[ 2L = 2500 \]\[ L = 1250 \]

Step 6 - Calculate the Corresponding Width

Substitute \(L = 1250\) back into the equation \(W = 2500 - L\): \[ W = 2500 - 1250 = 1250 \]

Step 7 - Compute the Area of the Large Rectangle and Each Pen

The area of the large rectangle is: \[ A = L \times W = 1250 \times 1250 = 1562500 \mathrm{ft^2} \]Since there are 4 equally sized pens, the area of each pen is: \[ \frac{A}{4} = \frac{1562500}{4} = 390625 \mathrm{ft^2} \]

Key Concepts

Rectangular Pasture

A rectangular pasture is often used in farming because of its simple shape, which makes it easy to enclose and divide into smaller sections. In this problem, the farmer wants a large rectangular pasture divided into four equally-sized pens. Each pen will contain different types of animals, so they don't mix.
To understand the maximum area each pen can have, we start by defining the overall dimensions of the large rectangle. Let’s denote the length by \(L\) and the width by \(W\). Each pen will then have dimensions \(\frac{L}{2} \times \frac{W}{2}\).
The total amount of available fencing is critical when setting up the pasture correctly, as this directly affects the dimensions that maximize the enclosed area for the animals.

Fencing Equation

The fencing equation helps us account for the total perimeter of the pasture and the internal divisions. In this exercise, the farmer has \(7500 \mathrm{ft}\) of fencing. The amount of fencing forms not only the outer rectangle but also the internal segments that partition the large rectangle into four smaller pens.

• The outer rectangle requires fencing for both lengths and widths: \(2L\) and \(2W\).
• Additionally, inner segments add three more pieces: \(W\) and \(L\).

We can combine these to form the total fencing equation: \[2L + 2W + W + L = 3L + 3W\]
Simplifying this equation by dividing by 3, we get: \(L + W = 2500\). This simplification makes it easier to handle the relationship between length and width as we proceed to maximize the area.

Critical Points

Critical points are essential in optimization problems; they help us find the maximum or minimum values. In this context, we're looking to maximize the area of the pasture. We express the width \(W\) in terms of length \(L\) using the simplified fencing equation: \(W = 2500 - L\).
The area of the large rectangle can be written in terms of \(L\) as: \[A = L \times W = L \times (2500 - L) = 2500L - L^2\]
To find the maximum area, we take the derivative of \(A\) with respect to \(L\): \(\frac{dA}{dL} = 2500 - 2L\). Setting this derivative equal to zero to identify critical points, we solve: \[2500 - 2L = 0 \] Solving for \(L\), we find \(L = 1250\).
Substituting \(L\) back into the equation for \(W\), we get \(W = 1250\). Thus, both length and width are equal, each \(1250 \mathrm{ft}\).
Finally, the area of the large rectangle is \(1250 \times 1250 = 1562500 \mathrm{ft^2}\). Dividing this by four, each pen has an area of \(390625 \mathrm{ft^2}\). Critical points help pinpoint the exact dimensions that yield this maximum area for optimal pasture design.