Maximize Area Rectangular Pen Using Quadratic Equations
In the world of practical mathematics, farmers often face optimization challenges that require a blend of geometric understanding and algebraic application. One such common challenge involves maximizing the area of a rectangular pen using a limited amount of fencing. This scenario not only tests the farmer's ability to calculate but also their understanding of how mathematical equations can model real-world constraints and outcomes. This article delves into a classic problem: a farmer with 100 meters of fencing aiming to create the largest possible rectangular enclosure. We will explore how to derive the quadratic equation that represents the area of the pen in terms of its width, and how this equation can be used to solve the farmer's optimization problem. The process involves translating the physical constraints of the fencing and the geometry of the rectangle into a mathematical model, allowing us to analyze the relationship between the pen's dimensions and its area. Understanding this relationship is crucial for anyone looking to apply mathematical principles to practical scenarios, whether in agriculture, construction, or any field involving resource optimization.
Defining the Problem: Perimeter and Area
The core of the problem lies in understanding the relationship between the perimeter and the area of a rectangle. The perimeter, in this case, is the total length of the fencing, which is a fixed 100 meters. This fencing will form the four sides of the rectangular pen. If we denote the width of the pen as w and the length as l, the perimeter P can be expressed as:
P = 2l + 2w
Since the farmer has 100 meters of fencing, we can write:
100 = 2l + 2w
The area A of the rectangular pen, which the farmer wants to maximize, is given by:
A = l * w*
These two equations form the foundation of our problem. The first equation represents the constraint (the limited fencing), and the second equation represents the objective (maximizing the area). To find the quadratic equation that expresses the area in terms of the width, we need to manipulate these equations to eliminate the length l and express the area A solely as a function of w. This involves algebraic substitution and rearrangement, which will lead us to a quadratic expression that we can analyze to find the maximum possible area.
Deriving the Quadratic Equation: A Step-by-Step Approach
To derive the quadratic equation that represents the area (A) of the pen in terms of its width (w), we need to express the length (l) in terms of w using the perimeter equation. Starting with the perimeter equation:
100 = 2l + 2w
We can simplify this by dividing both sides by 2:
50 = l + w
Now, we solve for l:
l = 50 - w
This equation tells us how the length of the pen is related to its width, given the constraint of 100 meters of fencing. Next, we substitute this expression for l into the area equation:
A = l * w*
A = (50 - w) * w*
Expanding this, we get:
A = 50w - w^2
This is the quadratic equation that expresses the area A of the pen as a function of its width w. The equation is in the form of a quadratic function, which is a polynomial function of degree two. The negative coefficient of the w^2 term indicates that the parabola opens downwards, meaning it has a maximum point. This maximum point corresponds to the maximum area that the farmer can enclose with the given amount of fencing.
Analyzing the Quadratic Equation: Understanding Maximum Area
Now that we have the quadratic equation A = 50w - w^2, we can analyze it to understand how the area of the pen changes with different widths. This equation is a parabola, and its graph would show the relationship between the width w and the area A. The vertex of this parabola represents the maximum area that can be enclosed.
To find the vertex, we can use the formula for the x-coordinate (in this case, the w-coordinate) of the vertex of a parabola given by the equation A = aw^2 + bw + c, which is:
w = -b / (2a)
In our equation, A = -w^2 + 50w, a = -1 and b = 50. So,
w = -50 / (2 * -1)
w = 25
This means that the width that maximizes the area is 25 meters. To find the maximum area, we substitute this value of w back into the area equation:
A = 50 * 25 - 25^2
A = 1250 - 625
A = 625
Therefore, the maximum area that the farmer can enclose is 625 square meters. This occurs when the width of the pen is 25 meters. To find the corresponding length, we use the equation l = 50 - w:
l = 50 - 25
l = 25
Thus, the length is also 25 meters. This tells us that the rectangle that maximizes the area is actually a square, with sides of 25 meters each.
Choosing the Correct Quadratic Equation: Options Analysis
In the original problem, we were given four options for the quadratic equation representing the area. Let's revisit those options in light of our derived equation:
A. A(w) = w^2 - 50w B. A(w) = w^2 - 100w C. A(w) = 50w - w^2 D. A(w) = 100w - w^2
Comparing these options with the equation we derived, A = 50w - w^2, we can see that option C matches our result. The other options either have the wrong signs or incorrect coefficients. Option A and B have a positive w^2 term, which would indicate a parabola that opens upwards, implying a minimum area rather than a maximum. Option D has the correct signs but the wrong coefficient for the w term. Therefore, the correct quadratic equation that gives the area of the pen, given its width, is:
C. A(w) = 50w - w^2
Conclusion: Practical Application of Quadratic Functions
This problem demonstrates a practical application of quadratic functions in optimizing real-world scenarios. By understanding the relationship between the perimeter and area of a rectangle, and by using algebraic manipulation to derive a quadratic equation, we were able to determine the dimensions that maximize the area of a rectangular pen with a fixed amount of fencing. The process involved several key steps: defining the problem, expressing the perimeter and area in terms of width and length, deriving the quadratic equation by substitution, analyzing the equation to find the maximum area, and choosing the correct equation from the given options.
The solution highlights the importance of mathematical modeling in problem-solving. Quadratic equations, with their parabolic nature, are particularly useful for optimization problems where there is a maximum or minimum value to be found. This type of problem-solving approach is not limited to agriculture; it can be applied in various fields, including engineering, economics, and computer science, where optimization is a critical consideration. Whether it's maximizing profit, minimizing cost, or optimizing resource allocation, the principles of quadratic functions and their applications remain a valuable tool in the arsenal of any problem solver.
