Quadratic Equation And Maximum Area Problem Solving A Farmer's Fencing Challenge
In this engaging exploration of quadratic equations, we delve into a practical problem faced by a farmer with 100 meters of fencing. The farmer aims to enclose a rectangular pen, and our goal is to determine the quadratic equation that represents the area (A) of the pen in relation to its width (w). Furthermore, we will unravel the process of finding the greatest rectangular area that the farmer can achieve with the given fencing. This involves understanding how the dimensions of the rectangle influence the enclosed area and how quadratic equations can be used to model and optimize such scenarios. This article aims to provide a comprehensive understanding of the problem, guiding you through the steps to derive the correct quadratic equation and calculate the maximum area. By exploring the relationship between the dimensions of the rectangular pen and its area, we'll gain valuable insights into the application of mathematical concepts in real-world situations. The farmer's challenge serves as a compelling example of how quadratic equations can be employed to solve optimization problems, making it a fundamental concept in mathematics and its applications.
Problem Statement
The core of our problem lies in the farmer's desire to maximize the area of a rectangular pen using a fixed length of fencing. The farmer has 100 meters of fencing at their disposal, and the challenge is to find the dimensions—width and length—that will result in the largest possible enclosed area. This is a classic optimization problem that can be effectively tackled using the principles of algebra and quadratic equations. To begin, we need to establish the relationship between the perimeter of the rectangle (which is the total length of the fencing) and its dimensions. The perimeter (P) of a rectangle is given by the formula P = 2l + 2w, where l represents the length and w represents the width. In this scenario, the perimeter is equal to the total fencing available, which is 100 meters. Therefore, we can express this relationship as 100 = 2l + 2w. From this equation, we can derive an expression for the length (l) in terms of the width (w), which will be crucial in formulating the quadratic equation for the area. The next step involves expressing the area (A) of the rectangle as a function of its width. The area of a rectangle is calculated by multiplying its length and width, A = l * w. By substituting the expression for l that we derived from the perimeter equation, we will obtain a quadratic equation that represents the area in terms of the width. This quadratic equation will then be the key to determining the maximum possible area and the corresponding dimensions of the rectangular pen. Understanding the problem statement thoroughly is essential for navigating the solution process effectively and appreciating the practical implications of the mathematical concepts involved.
Deriving the Quadratic Equation
The journey to finding the perfect pen area begins with understanding the connection between the fence's length and the pen's dimensions. We know the perimeter (P) of a rectangle is 100 meters, which can be expressed as P = 2l + 2w, where l is the length and w is the width. Let's translate this into an equation: 100 = 2l + 2w. Our mission is to find a quadratic equation that represents the area (A) of the pen in terms of its width (w). To do this, we must first express the length l in terms of the width w. We can simplify the perimeter equation by dividing both sides by 2, giving us 50 = l + w. Now, we can isolate l by subtracting w from both sides, resulting in l = 50 - w. This equation tells us exactly how the length changes with the width, keeping the total fencing at 100 meters. Next, we recall that the area (A) of a rectangle is calculated by multiplying its length and width, so A = l * w. We've already found an expression for l in terms of w, so we can substitute l = 50 - w into the area equation. This gives us A = (50 - w) * w. Expanding this equation, we get A = 50w - w². This is the quadratic equation we've been searching for! It beautifully captures how the area of the rectangular pen changes as we vary the width, given the constraint of 100 meters of fencing. The equation A(w) = 50w - w² is a crucial step forward, allowing us to analyze the relationship between the width and area and, ultimately, to find the maximum possible area. This equation is the key to understanding the optimization problem at hand.
Analyzing the Quadratic Equation
Now that we have derived the quadratic equation A(w) = 50w - w², the next crucial step is to analyze it to understand its implications for the farmer's pen. This equation is in the form of a quadratic function, which, when graphed, forms a parabola. The shape of the parabola is critical to our understanding because it will help us identify the maximum area that the farmer can enclose with the given fencing. The general form of a quadratic equation is ax² + bx + c, where a, b, and c are constants. In our case, the equation A(w) = 50w - w² can be rewritten as A(w) = -w² + 50w, which means a = -1, b = 50, and c = 0. The coefficient a plays a significant role in determining the shape of the parabola. Since a is negative (-1 in our case), the parabola opens downwards. This means that the graph has a maximum point, which is the vertex of the parabola. The vertex represents the highest point on the graph and corresponds to the maximum area that can be enclosed. To find the width (w) that corresponds to this maximum area, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex of a parabola given by the equation ax² + bx + c is -b/(2a). In our equation, this translates to w = -50/(2*(-1)) = 25. So, the width that maximizes the area is 25 meters. Understanding the properties of the quadratic equation and the significance of the parabola's vertex is essential for solving the optimization problem. The fact that the parabola opens downwards and has a maximum point is what allows us to find the dimensions that will yield the greatest possible area for the farmer's pen. This analysis provides the foundation for the subsequent steps in calculating the maximum area and determining the optimal length of the pen.
Finding the Greatest Rectangular Area
With the quadratic equation A(w) = 50w - w² firmly in hand, we've already pinpointed the width that promises the largest area: 25 meters. Now, let's calculate the maximum area itself. We'll simply plug this width value back into our area equation. So, A(25) = 50(25) - (25)² = 1250 - 625 = 625 square meters. This is the largest possible area the farmer can enclose with the 100 meters of fencing! But we're not quite done yet. We also need to determine the length of the pen when the width is 25 meters. Remember, we derived the equation l = 50 - w earlier. Substituting w = 25 meters into this equation, we get l = 50 - 25 = 25 meters. Interestingly, we find that the length is also 25 meters. This reveals a crucial insight: the rectangular pen that encloses the maximum area is actually a square! A square, with its equal sides, maximizes the area for a given perimeter. This is a fundamental principle in geometry and optimization. The dimensions of the pen that yield the greatest area are therefore 25 meters by 25 meters, resulting in an area of 625 square meters. This solution highlights the power of quadratic equations in solving practical problems. By understanding the relationship between the dimensions of the rectangle and its area, we've successfully optimized the pen's design to maximize the enclosed space. The farmer can now confidently build the pen, knowing they've made the most efficient use of their fencing. This process underscores the importance of mathematical modeling in real-world applications, allowing us to make informed decisions and achieve the best possible outcomes.
Practical Implications and Conclusion
In conclusion, this exercise with the farmer's fencing and the rectangular pen vividly demonstrates the practical applications of quadratic equations. We started with a real-world problem—maximizing the area of a pen with a limited amount of fencing—and used mathematical principles to arrive at an optimal solution. The derived quadratic equation, A(w) = 50w - w², became our key tool in understanding the relationship between the width of the pen and its area. By analyzing this equation, we discovered that the maximum area is achieved when the pen is a square, with dimensions of 25 meters by 25 meters, enclosing an area of 625 square meters. This solution not only provides a specific answer to the farmer's problem but also highlights a broader principle: for a given perimeter, a square will always maximize the enclosed area among all rectangles. This has practical implications in various fields, from construction and design to agriculture and resource management. Understanding how to optimize area with limited resources is a valuable skill in many professions. Furthermore, this problem underscores the importance of mathematical modeling in problem-solving. By translating a real-world scenario into a mathematical equation, we can use the tools of algebra and calculus to find solutions that might not be immediately apparent. The process of deriving and analyzing the quadratic equation also reinforces fundamental mathematical concepts such as functions, parabolas, and optimization. This exercise serves as an excellent example of how mathematical knowledge can be applied to solve practical problems and make informed decisions. The farmer, armed with this understanding, can now confidently build the pen, knowing they've maximized the enclosed space for their livestock or crops. In essence, this problem illustrates the power of mathematics to not only explain the world around us but also to improve it.
Therefore, the quadratic equation that gives the area (A) of the pen, given its width (w), is A(w) = 50w - w². The greatest rectangular area that the farmer can enclose is 625 square meters.
