FGV 2017 Maximizing Area A Farmer's Rectangular Fence Problem Discussion

Introduction to the Farmer's Fence Problem

In this comprehensive discussion, we delve into a classic optimization problem that has significant real-world applications: maximizing the area enclosed by a rectangular fence given a fixed perimeter. This problem, often encountered in introductory calculus and optimization courses, provides a fantastic opportunity to explore the relationship between perimeter, area, and the use of mathematical principles to solve practical scenarios. Specifically, we will dissect a problem presented in the FGV 2017 exam, which challenges us to determine the dimensions of a rectangular fence that will yield the largest possible area for a farmer, considering a constraint on the total length of fencing material available. This exploration will not only reinforce fundamental mathematical concepts but also highlight the importance of optimization in everyday decision-making, from agriculture to urban planning. Understanding how to maximize the area within a given constraint is crucial for efficient resource utilization and cost-effectiveness. We'll embark on a step-by-step journey, from setting up the problem using algebraic equations to employing calculus techniques to arrive at the optimal solution. By the end of this discussion, you'll have a robust understanding of how to tackle similar optimization problems, equipping you with valuable skills for both academic pursuits and real-life challenges. The Farmer's Fence Problem is not just a theoretical exercise; it’s a gateway to appreciating the power of mathematics in shaping our world and finding the most efficient solutions to practical constraints. It underscores the essence of optimization: achieving the most desirable outcome within given limitations. So, let's roll up our sleeves and get ready to cultivate our mathematical prowess as we unravel this intriguing problem.

Problem Statement: FGV 2017

Let's precisely define the problem we're tackling. Imagine a farmer who has a limited amount of fencing material and wants to enclose a rectangular area for his livestock. The challenge lies in determining the dimensions of the rectangle – the length and the width – that will yield the maximum possible area, given the constraint of the fixed fencing length. This is the core of the FGV 2017 problem. The problem statement, as presented in the FGV 2017 exam, provides specific numerical constraints. For the purpose of this discussion, let's assume the farmer has 100 meters of fencing material. This constraint is crucial because it dictates the perimeter of the rectangular enclosure. The perimeter, being the total length of the fence, is the sum of all the sides of the rectangle. The farmer's goal is simple yet impactful: to create the largest possible grazing area for his animals using the available resources efficiently. To solve this, we must translate this real-world scenario into a mathematical model. We'll introduce variables to represent the length and width of the rectangle and formulate equations that capture the relationship between these dimensions, the perimeter, and the area. The crux of the problem lies in understanding that while many different rectangles can be formed with the same perimeter, they will not all have the same area. Some rectangles will be long and narrow, while others will be closer to a square shape. Our objective is to find the specific dimensions that strike the perfect balance, maximizing the enclosed area. This involves a blend of algebraic manipulation and calculus techniques, ultimately leading us to the optimal solution. This FGV 2017 problem serves as an excellent illustration of how mathematical optimization can be applied to practical scenarios, emphasizing the importance of careful planning and resource management.

Setting Up the Mathematical Model

To effectively address the farmer's fencing dilemma, we must first translate the problem into a language mathematics understands. This involves setting up a mathematical model that accurately represents the scenario and the constraints. Let's start by defining our variables. Let 'l' represent the length of the rectangular fence and 'w' represent its width. These are the two dimensions we aim to determine to maximize the area. The area 'A' of a rectangle is given by the product of its length and width: A = l * w. This is the function we want to maximize. Next, we need to incorporate the constraint: the fixed amount of fencing material. As mentioned earlier, let's assume the farmer has 100 meters of fencing. The perimeter 'P' of a rectangle is given by the sum of all its sides, which is P = 2l + 2w. Since the farmer has 100 meters of fencing, we have the constraint equation: 2l + 2w = 100. This equation links the length and width, imposing a limitation on the possible values they can take. We now have two equations: one for the area we want to maximize (A = l * w) and one for the constraint (2l + 2w = 100). To solve this optimization problem, we need to express the area 'A' as a function of a single variable. We can achieve this by using the constraint equation to solve for one variable in terms of the other. For instance, we can solve for 'w' in terms of 'l': 2w = 100 - 2l, which simplifies to w = 50 - l. Now, we can substitute this expression for 'w' into the area equation: A = l * (50 - l). This gives us A(l) = 50l - l², a quadratic function representing the area as a function of the length. This is a crucial step because it allows us to use calculus techniques to find the maximum value of the area. We have successfully transformed a real-world problem into a mathematical model, setting the stage for the next step: finding the optimal dimensions.

Solving for Maximum Area Using Calculus

Now that we have our area function, A(l) = 50l - l², we can leverage the power of calculus to find the length 'l' that maximizes the area. The key concept here is that the maximum or minimum of a function occurs where its derivative is equal to zero. The derivative of a function represents its rate of change, and at the peak or trough of a curve, the rate of change is momentarily zero. So, our first step is to find the derivative of the area function with respect to 'l'. The derivative, denoted as A'(l), is calculated as follows: A'(l) = d(50l - l²)/dl = 50 - 2l. Next, we set the derivative equal to zero and solve for 'l': 50 - 2l = 0. This gives us 2l = 50, and therefore, l = 25 meters. This value of 'l' is a critical point, a potential location for a maximum or minimum of the area function. To confirm that this value corresponds to a maximum and not a minimum, we can use the second derivative test. The second derivative, A''(l), is the derivative of A'(l): A''(l) = d(50 - 2l)/dl = -2. Since the second derivative is negative, this indicates that the area function is concave down at l = 25, confirming that it is indeed a maximum. Now that we have the optimal length, l = 25 meters, we can find the corresponding width 'w' using the equation we derived earlier: w = 50 - l. Substituting l = 25, we get w = 50 - 25 = 25 meters. Thus, the dimensions that maximize the area are l = 25 meters and w = 25 meters. This means the rectangle that encloses the maximum area is a square. To find the maximum area itself, we substitute these values back into the area equation: A = l * w = 25 * 25 = 625 square meters. Therefore, the farmer can enclose a maximum area of 625 square meters with the 100 meters of fencing by creating a square enclosure with sides of 25 meters each. This elegant solution demonstrates the power of calculus in solving optimization problems.

Interpretation of the Solution

Our calculations have revealed that the dimensions that maximize the area enclosed by the farmer's fence are 25 meters for both the length and the width, resulting in a square shape. This leads to a maximum enclosed area of 625 square meters. But what does this solution tell us in practical terms? It highlights a fundamental principle in geometry and optimization: for a given perimeter, a square encloses the largest possible area among all rectangles. This isn't just a mathematical curiosity; it has practical implications in various fields. In agriculture, for instance, understanding this principle allows farmers to efficiently utilize their resources, ensuring they get the most grazing area for their livestock with a given amount of fencing. In urban planning, it can inform decisions about the shapes of parks and other recreational spaces, maximizing the usable area for the community. The solution also underscores the concept of diminishing returns. While increasing the perimeter (i.e., using more fencing) will undoubtedly increase the area, the shape of the enclosure plays a crucial role in how efficiently that perimeter is used. A long, narrow rectangle, while having the same perimeter as a square, will enclose a significantly smaller area. The square, by distributing the perimeter equally among its sides, achieves the optimal balance. Furthermore, this problem provides a tangible example of how mathematical models can be used to solve real-world challenges. By translating the farmer's fencing dilemma into algebraic equations and then employing calculus techniques, we were able to arrive at a precise and actionable solution. This process of modeling, solving, and interpreting is a cornerstone of applied mathematics and is used extensively in fields ranging from engineering to economics. In essence, the solution to the FGV 2017 problem is not just about finding numbers; it's about understanding the underlying principles of optimization and appreciating the power of mathematical reasoning in making informed decisions.

Alternative Approaches and Extensions

While we successfully used calculus to solve the farmer's fencing problem, it's worth noting that there are alternative approaches to arrive at the same solution. One such approach involves using the concept of completing the square, a technique from algebra. Recall our area function: A(l) = 50l - l². We can rewrite this quadratic expression by completing the square. This involves manipulating the expression to create a perfect square trinomial. To do this, we take half of the coefficient of the 'l' term (which is 50), square it (which gives us 625), and add and subtract it within the expression: A(l) = -(l² - 50l). Now, we add and subtract (50/2)² = 625 inside the parenthesis: A(l) = -(l² - 50l + 625 - 625). This allows us to rewrite the expression as: A(l) = -(l - 25)² + 625. This form of the equation reveals that the maximum area occurs when (l - 25)² is minimized, which happens when l = 25. The maximum area is then 625 square meters, confirming our calculus-based solution. This alternative approach provides a valuable perspective, demonstrating that the same problem can often be solved using different mathematical tools. Beyond the core problem, we can also consider extensions and variations. For instance, what if the farmer's enclosure is bounded by a river on one side, so fencing is only needed for the other three sides? This modifies the constraint equation, leading to a different optimal solution. Or, what if the farmer wants to divide the enclosure into multiple sections using additional fencing? This introduces new variables and constraints, making the problem more complex. Another extension could involve considering non-rectangular shapes. Would a circle, for example, enclose a larger area than a square with the same perimeter? Exploring these variations not only deepens our understanding of optimization but also highlights the adaptability of mathematical models to different scenarios. The farmer's fencing problem, while seemingly simple, serves as a springboard for exploring a wide range of mathematical concepts and applications.

Conclusion

In conclusion, the FGV 2017 farmer's fencing problem provides a compelling illustration of how mathematical optimization can be applied to a real-world scenario. By carefully setting up a mathematical model, using algebraic equations to represent the constraints and the objective function, and then employing calculus techniques (or alternative methods like completing the square), we were able to determine the dimensions of the rectangular fence that maximize the enclosed area. The solution, a square with sides of 25 meters, highlights the principle that for a given perimeter, a square encloses the largest possible area among all rectangles. This understanding has practical implications in various fields, from agriculture to urban planning, emphasizing the importance of efficient resource utilization. Furthermore, the problem serves as a valuable exercise in mathematical problem-solving, demonstrating the power of translating real-world scenarios into mathematical models and using analytical tools to arrive at optimal solutions. The extensions and variations we discussed further illustrate the versatility of these techniques and their applicability to a wide range of problems. The farmer's fencing problem is more than just a theoretical exercise; it's a testament to the power of mathematical thinking in making informed decisions and maximizing outcomes in the face of constraints. It reinforces the idea that mathematics is not just an abstract subject but a powerful tool for understanding and shaping the world around us. The ability to model, solve, and interpret optimization problems is a valuable skill, applicable in numerous contexts, making this classic problem a cornerstone of mathematical education.