Question

A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

Short answer

The largest area that can be enclosed is 500,000 square meters.

Step-by-step solution

Define the Variables

Let the length of the plot parallel to the highway be denoted as \( L \), and the width perpendicular to the highway be denoted as \( W \).

Set Up the Constraint

Since the farmer does not fence the side along the highway, the total amount of fencing used will be for two widths and one length. Therefore, the constraint equation is: \[ 2W + L = 2000 \]

Express Length in Terms of Width

Solve the constraint equation for \( L \): \[ L = 2000 - 2W \]

Write the Area Formula

The area \( A \) of the rectangle can be expressed as the product of length and width: \[ A = L \cdot W \] Substitute the expression for \( L \) into the area formula: \[ A = (2000 - 2W) \cdot W \]

Simplify and Differentiate

Simplify the area formula into a quadratic equation: \[ A = 2000W - 2W^2 \] To find the maximum area, take the derivative of \( A \) with respect to \( W \) and set it to zero: \[ \frac{dA}{dW} = 2000 - 4W = 0 \]

Solve for Width

Solve the equation for \( W \): \[ 4W = 2000 \] \[ W = 500 \]

Find the Length

Substitute \( W = 500 \) back into the constraint equation to find \( L \): \[ L = 2000 - 2(500) = 1000 \]

Calculate the Maximum Area

Substitute \( W = 500 \) and \( L = 1000 \) back into the area formula: \[ A = 1000 \cdot 500 = 500000 \]

Key concepts

Maximum Area Problem

In this problem, we are tasked with maximizing the area of a rectangular plot using a given amount of fencing. A careful approach is needed when determining the dimensions to achieve the largest area. This kind of problem is common in optimization, where you need to find the highest or lowest value of a certain quantity under given constraints.

Here, the farmer uses fencing to enclose three sides of the rectangular plot, with one side bordering the highway. Our target is to make the best use of the available 2000 meters of fencing to achieve the maximum enclosed area. Understanding how to set up and solve this type of problem can be extremely useful in various practical applications, such as agriculture, landscaping, and even in optimizing resources for projects.

Quadratic Equations

During the optimization process, we transform the problem into a quadratic equation. Quadratic equations are polynomial equations of the form: a(x) = ax^2 + bx + c In this problem, the area formula becomes a quadratic equation after substituting the expressions for length and width. Specifically:

A = 2000W - 2W^2

This formula represents a parabola, and our goal is to find the maximum value of this parabola. The vertex of a downward-opening parabola, which in this case occurs at the value of W that makes the first derivative zero, gives us the maximum area.

Derivative Applications

Taking derivatives is a key step in finding the maximum area. The derivative of a function helps determine its critical points—where the value could be a maximum or minimum. In this problem, we take the first derivative of the area function:
\[\frac{dA}{dW} = 2000 - 4W\]

This derivative tells us how the area changes with respect to the width. Setting the derivative equal to zero finds the point at which the area is maximized: \[2000 - 4W = 0\], leading to \[W = 500\], which helps in finding the critical width.

By solving for W, we then determine the dimensions that maximize the area. Differentiation thus offers a powerful tool in the optimization process, especially in calculus problems involving maximum and minimum values.

Rectangular Plot

Understanding the structure of the rectangular plot is crucial. This problem involves a rectangle, where we have three sides fenced, and one side left open along the highway. Let's break down the steps:

• Length (L): Parallel to the highway.
• Width (W): Perpendicular to the highway.

Given that we are using 2000 meters of fencing, the constraints help in formulating equations to express one dimension in terms of another. The interplay between length and width under the fencing constraint is pivotal in finding the optimal dimensions for the largest possible area.

Constraint Equations

In optimization problems like this one, constraints form the backbone for deriving solutions. A constraint equation represents the limits within which we must work. For our problem, the constraint equation is:

\[2W + L = 2000\]

This equation arises because there are two widths and one length to be fenced. Solving this for one variable in terms of the other (here L in terms of W) simplifies the problem:

\[ L = 2000 - 2W \]

By expressing the length in terms of width, we can substitute it back into the area formula, leading to a single-variable quadratic equation. This simplification allows us to use calculus tools like differentiation to find the maximum area efficiently. Constraints help shape the feasible solutions and guide the optimization process.