Problem 75 - Precalculus Mathematics for Calculus

Question

Fencing a Horse Corral Carol has \(2400 \mathrm{ft}\) of fencing to fence in a rectangular horse corral. (a) Find a function that models the area of the corral in terms of the width \(x\) of the corral. (b) Find the dimensions of the rectangle that maximize the area of the corral.

Step-by-Step Solution

Answer

The area is maximized for a square with dimensions 600 ft by 600 ft.

Step 1: Understand the problem

We need to create a function that represents the area of a rectangular corral using a given amount of fencing, and then find the dimensions that maximize this area.

Step 2: Set up the perimeter equation

The perimeter of a rectangle is given by the equation: \( P = 2x + 2y \), where \( x \) is the width and \( y \) is the length. We know that the total perimeter is 2400 ft, so we have: \( 2x + 2y = 2400 \).

Step 3: Simplify the perimeter equation to express y in terms of x

Divide the perimeter equation by 2: \( x + y = 1200 \). Solve for \( y \): \( y = 1200 - x \).

Step 4: Express the area function in terms of x

The area \( A \) of the rectangle is \( A = x \cdot y \). Substitute \( y \) from the simplified perimeter equation: \( A = x(1200 - x) \). Simplify the expression to get: \( A(x) = 1200x - x^2 \). This is the area function in terms of \( x \).

Step 5: Find the value of x that maximizes the area using derivatives

To find the maximum area, take the derivative of \( A(x) \) with respect to \( x \): \( A'(x) = 1200 - 2x \). Set the derivative equal to zero to find critical points: \( 1200 - 2x = 0 \). Solve for \( x \): \( x = 600 \).

Step 6: Use the value of x to find y

Substitute \( x = 600 \) back into the equation \( y = 1200 - x \) to find \( y \): \( y = 1200 - 600 = 600 \). Therefore, the dimensions that maximize the area are 600 ft by 600 ft.

Key Concepts

Area Maximization

When we talk about area maximization, particularly for Carol's horse corral, we are aiming to enclose the largest possible rectangular area with a fixed amount of fencing. The concept of maximizing area is crucial in many real-world applications where limited resources need to be optimally used.
To find the maximum area, we start by considering a function that expresses the area based on one of the dimensions (in this case, the width \(x\)). This requires understanding the relationship between the dimensions and the constraint given by the perimeter. Once the area function is established, calculus tools, such as derivatives, are used to find the point where this function reaches its maximum. In our example, it turned out the maximum area was achieved when both the width and length were equal, leading to the optimal use of the fencing available.

Rectangular Dimensions

For the problem of finding dimensions that maximize the area with a given perimeter, we first need to establish the relationship between width \(x\) and length \(y\). The total perimeter is simply twice the sum of its width and length:
\[ 2x + 2y = 2400 \]
Upon simplifying, we relate these dimensions with:\
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\( y = 1200 - x \)

This relationship was used to frame the area of the rectangle in terms of \(x\). In the pursuit of maximum area, recognizing how these dimensions interplay is fundamental.
In our solution, noticing that when \(x\) and \(y\) are equal, the area function attained its peak pointed towards a common mathematical insight. The unique features of a square (having all equivalent sides) come naturally into play when the perimeter is split evenly among width and length.

Derivative Application

Derivatives are powerful tools in calculus that help us find maximum or minimum values of functions, like in Carol's corral fencing problem. The derivative of a function gives us the rate at which the function value changes as its input changes.
The area function for the corral was \(A(x) = 1200x - x^2\). To find where the area is maximized, we take the derivative with respect to \(x\):
\[ A'(x) = 1200 - 2x \]
By setting this derivative to zero, \(1200 - 2x = 0\), we identify the critical point that potentially offers the maximum area.
After solving the derivative equation, we found \(x = 600\), pinpointing where the rate of change of the area shifts from increasing to decreasing, thereby achieving the maximum. This derivative test indicates not only where a maximum is but helps provide confirmation that the maximum area uses all 2400 ft of fencing most efficiently.