Quadratic Equation For Rectangular Pen Area Given Fencing Length

In the realm of practical mathematics, optimizing area within given constraints is a common challenge. This article delves into a classic problem: A farmer with 100 meters of fencing aims to enclose a rectangular pen. Our goal is to determine the quadratic equation that represents the pen's area (A) in relation to its width (w). This problem isn't just a mathematical exercise; it's a fundamental concept in optimization, with applications spanning from agriculture and engineering to economics and urban planning. Understanding how to express real-world scenarios with quadratic equations allows us to make informed decisions about resource allocation and maximizing desired outcomes.

To begin, let's clarify the farmer's situation. He possesses 100 meters of fencing, which will serve as the perimeter of the rectangular pen. A rectangle's perimeter is defined as twice the sum of its length (l) and width (w), mathematically expressed as 2l + 2w. Therefore, we know that 2l + 2w = 100. The area (A) of a rectangle, on the other hand, is calculated by multiplying its length and width, A = l * w. Our challenge is to express A as a function of w alone, resulting in a quadratic equation. This requires us to eliminate the length (l) variable from the area equation using the perimeter constraint. By doing so, we'll have a clear relationship between the pen's width and its potential area, allowing us to explore how different widths affect the enclosed space. This initial setup is crucial for translating the word problem into a mathematical model, which is the foundation for solving optimization problems.

To derive the quadratic equation that gives the area (A) of the pen, given its width (w), we first need to express the length (l) in terms of the width using the perimeter constraint. We know that the perimeter is 100 meters, and the formula for the perimeter of a rectangle is 2l + 2w = 100. We can solve for l as follows:

1. Divide the entire equation by 2: l + w = 50
2. Subtract w from both sides: l = 50 - w

Now that we have the length expressed in terms of the width, we can substitute it into the area equation, A = l * w. Replacing l with (50 - w), we get:

A = (50 - w) * w

Expanding this equation, we obtain:

A = 50w - w^2

This is a quadratic equation in the form A(w) = -w^2 + 50w, which represents the area of the pen as a function of its width. The negative coefficient of the w^2 term indicates that the parabola opens downward, meaning there's a maximum area that can be achieved. This equation allows us to explore how different widths affect the area and ultimately find the width that maximizes the pen's enclosed space. The derived quadratic equation is a crucial tool for optimizing the pen's dimensions, demonstrating the practical application of mathematical modeling in real-world scenarios.

Now, let's compare the derived quadratic equation, A(w) = 50w - w^2, with the options provided. The options are:

A. A(w) = 100w - w^2 B. A(w) = w^2 - 50w C. A(w) = w^2 - 100w D. A(w) = 50w - w^2

By direct comparison, we can see that option D, A(w) = 50w - w^2, matches the equation we derived. The other options have different coefficients or signs, indicating they do not correctly represent the relationship between the pen's width and area given the 100-meter fencing constraint. Option A has a coefficient of 100 for the w term, which is incorrect based on our perimeter calculation. Options B and C have a positive coefficient for the w^2 term, which would imply the parabola opens upwards, indicating a minimum area rather than a maximum, which doesn't align with the problem's context. Therefore, a careful comparison with the derived equation confirms that option D is the correct representation of the pen's area as a function of its width. This analytical step highlights the importance of verifying the solution against the original problem and the derived mathematical model.

In conclusion, the quadratic equation that accurately represents the area (A) of the rectangular pen, given its width (w) and a 100-meter fencing constraint, is A(w) = 50w - w^2. This equation was derived by first establishing the relationship between the pen's length and width using the perimeter formula and then substituting that relationship into the area formula. Comparing the derived equation with the provided options, we confirmed that option D is the correct answer. This problem demonstrates the practical application of quadratic equations in optimization scenarios. Understanding how to formulate and solve such problems is crucial in various fields, including agriculture, engineering, and resource management. The ability to translate real-world constraints into mathematical models and analyze them to find optimal solutions is a valuable skill. By using the derived quadratic equation, the farmer can determine the width that maximizes the pen's area, making the most efficient use of the available fencing. This example serves as a reminder of the power of mathematics in solving everyday challenges and optimizing resource utilization.