Problem 5 - Calculus Early Transcendentals

Question

A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy-duty fencing selling for \(\$ 3\) a foot, while the remaining two sides will use standard fencing selling for \(\$ 2\) a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of \(\$ 6000 ?\)

Step-by-Step Solution

Answer

The dimensions of the greatest area are 500 feet by 750 feet.

Step 1: Define Variables and Equations

Let's denote the length of the plot as \( x \) feet and the width as \( y \) feet. Two opposite sides use heavy-duty fencing, so they have a total cost of \( 3 \times 2x \), while the other two sides use standard fencing, costing \( 2 \times 2y \). The total cost of the fencing is: \( 6x + 4y = 6000 \). The area to be maximized is \( A = xy \).

Step 2: Solve for One Variable

Rearrange the cost equation, \( 6x + 4y = 6000 \), for one variable: \( y = \frac{6000 - 6x}{4} = 1500 - 1.5x \).

Step 3: Substitute into Area Equation

Substitute \( y = 1500 - 1.5x \) into the area equation: \( A = x(1500 - 1.5x) = 1500x - 1.5x^2 \).

Step 4: Differentiate the Area Equation

Differentiate the area function with respect to \( x \): \( \frac{dA}{dx} = 1500 - 3x \).

Step 5: Find Critical Points

Set the derivative equal to zero to find critical points: \( 1500 - 3x = 0 \) gives \( x = 500 \).

Step 6: Determine Dimensions with the Critical Point

Substitute \( x = 500 \) back into the equation for \( y \): \( y = 1500 - 1.5 \times 500 = 750 \). Thus, the dimensions are 500 feet by 750 feet.

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand how things change. It is often used in optimization problems where we want to find maximum or minimum values of functions. In the problem of finding the optimal dimensions of a fenced rectangular plot, calculus is used to maximize the area of the plot while considering the cost constraints.
To achieve this, we use derivatives, which are fundamental concepts in calculus. Derivatives help us find the rate at which a function (in this case, the area of the plot) changes as we vary one of the dimensions.
The critical points, found by setting the derivative to zero, tell us where the function might have maximum or minimum values. By identifying these points, we can determine the dimensions that provide the largest possible area within the given budget constraints.

Area Maximization

Area maximization is the goal of finding the largest possible area that can be enclosed given certain constraints. In this exercise, the largest area of a rectangular plot is sought under a fixed cost constraint for fencing.
The expression for the area of a rectangle is given by the equation \(A = x imes y\), where \(x\) and \(y\) are the length and width of the rectangle, respectively.
By using the constraints given by the cost of fencing, we can express one of the dimensions in terms of the other. This allows us to rewrite the area equation in terms of a single variable, making it easier to work with calculus to find maximum values. We derived the equation \(A = 1500x - 1.5x^2 \) and then used differentiation to find the point where the area is maximized.

Cost Constraints

Cost constraints are limitations set by budget restrictions, dictating that the total cost cannot exceed a specific amount. They play a crucial role in optimization problems like the one presented in this exercise.
By defining cost constraints, such as the cost per foot for different types of fencing, we translate these into mathematical equations that bound our problem.
In our case, the total cost equation \(6x + 4y = 6000\) arises because two sides of the rectangle use heavy-duty fencing at \(\\(3\) per foot, and the other two sides use standard fencing at \(\\)2\) per foot. Solving this equation allows us to express one dimension, \(y\), in terms of the other, \(x\). This connection between cost and dimensions is key to utilizing the budget efficiently while maximizing the area.