Maximizing Rectangular Pen Area A Farmer's Optimization Problem
As farmers seek to optimize their land use, maximizing the area they can enclose with a given amount of fencing is a common problem. This article delves into a specific scenario where a farmer aims to enclose a rectangular pen, with the area (A) described by a quadratic equation in terms of the width (w): A(w) = 50w - w^2. We will explore how to determine the greatest rectangular area the farmer can enclose with 100 meters of fencing.
Understanding the Problem
The core challenge lies in understanding the relationship between the dimensions of the rectangular pen (width and length) and its area, given a fixed perimeter (the fencing). The quadratic equation A(w) = 50w - w^2 provides a mathematical model for this relationship. Our objective is to find the width (w) that yields the maximum area (A) while adhering to the constraint of 100 meters of fencing.
To solve this optimization problem, we will employ techniques from algebra and calculus. First, we need to establish the connection between the width (w), length (l), and the perimeter of the rectangle. The perimeter is given by P = 2w + 2l. Since the farmer has 100 meters of fencing, we have 100 = 2w + 2l. This equation allows us to express the length (l) in terms of the width (w) and vice versa. Next, we can substitute this expression for length into the area formula, which is A = wl, to obtain the area as a function of a single variable, the width (w). This will give us the same quadratic equation provided, A(w) = 50w - w^2, but with a clearer understanding of its derivation from the geometric constraints of the problem.
Now, the problem is reduced to finding the maximum value of the quadratic function A(w). Quadratic functions have a parabolic shape, and their maximum (or minimum) value occurs at the vertex of the parabola. There are several ways to find the vertex. One method is to complete the square, which transforms the quadratic equation into vertex form, making the coordinates of the vertex readily apparent. Another approach involves using calculus. We can find the critical points of the function by taking its derivative with respect to w and setting it equal to zero. These critical points represent potential maxima or minima. The second derivative test can then be used to determine whether a critical point corresponds to a maximum or a minimum. In this case, since the coefficient of the w^2 term in A(w) is negative, the parabola opens downwards, indicating that the vertex represents a maximum. Once we find the width (w) that maximizes the area, we can substitute it back into the equation for length (l) in terms of w to find the corresponding length. Finally, we can calculate the maximum area by multiplying the optimal width and length.
By carefully applying these steps, we can determine the greatest rectangular area the farmer can enclose with 100 meters of fencing. This problem exemplifies the practical applications of mathematical concepts in real-world scenarios, highlighting the importance of optimization techniques in various fields.
Setting up the Perimeter Equation
The key to solving this optimization problem lies in understanding the constraint imposed by the 100 meters of fencing. The fencing represents the perimeter of the rectangular pen. We know that the perimeter (P) of a rectangle is given by the formula:
where:
- w is the width of the rectangle
- l is the length of the rectangle
In this case, the farmer has 100 meters of fencing, so we can write the equation:
This equation establishes a relationship between the width and length of the pen. We can simplify this equation by dividing both sides by 2:
Now, we can express the length (l) in terms of the width (w):
This equation is crucial because it allows us to relate the length to the width, enabling us to express the area of the pen as a function of a single variable, the width (w). This is a fundamental step in optimization problems, as it allows us to use techniques from calculus or algebra to find the maximum or minimum value of a function.
The equation l = 50 - w tells us that as the width increases, the length decreases, and vice versa, while maintaining a constant perimeter of 100 meters. This inverse relationship between width and length is what creates the possibility of maximizing the area. If the width is very small, the length will be large, but the area will be small because it's the product of width and length. Similarly, if the width is very large, the length will be small, again resulting in a small area. There must be an optimal balance between width and length that maximizes the area.
By substituting this expression for l into the area formula (A = wl), we will obtain a quadratic equation in terms of w, which is given in the problem statement as A(w) = 50w - w^2. This equation is a parabola, and its maximum value occurs at its vertex. Finding the vertex will give us the optimal width that maximizes the area of the pen. We can use various methods to find the vertex, such as completing the square, using the vertex formula, or employing calculus techniques. Each of these methods will lead us to the same solution, the width that maximizes the area given the constraint of 100 meters of fencing. The next step in the solution process is to substitute the expression l = 50 - w into the area formula and then proceed with finding the maximum value of the resulting quadratic function.
This process of setting up the perimeter equation and expressing one variable in terms of the other is a common strategy in optimization problems involving geometric shapes. It allows us to reduce the number of variables and express the quantity to be optimized (in this case, the area) as a function of a single variable, making it easier to apply optimization techniques.
Expressing Area as a Function of Width
Having established the relationship between the length (l) and width (w) of the rectangular pen using the perimeter constraint, we can now express the area (A) solely as a function of the width (w). This is a crucial step in solving the optimization problem because it allows us to use techniques for finding the maximum value of a function of a single variable.
Recall that the area of a rectangle is given by:
We previously derived the equation relating length and width from the perimeter constraint:
Now, we can substitute this expression for l into the area formula:
Expanding this expression, we get:
This is the quadratic equation given in the problem statement. It represents the area of the rectangular pen as a function of its width. The equation is a parabola that opens downwards because the coefficient of the w^2 term is negative. This means that the parabola has a maximum point, which corresponds to the maximum area that can be enclosed by the pen. Our goal is to find the width (w) that corresponds to this maximum point.
The quadratic function A(w) = 50w - w^2 provides a complete mathematical representation of the problem. It encapsulates both the geometry of the rectangle and the constraint imposed by the fixed amount of fencing. The graph of this function is a parabola, and the maximum area corresponds to the vertex of the parabola. To find the vertex, we can use several methods. One method is to complete the square, which transforms the quadratic equation into vertex form, making the coordinates of the vertex readily apparent. Another approach is to use the vertex formula, which states that the x-coordinate (in this case, the w-coordinate) of the vertex is given by -b/(2a), where a and b are the coefficients of the quadratic equation. A third method involves calculus: we can find the critical points of the function by taking its derivative and setting it equal to zero. These critical points represent potential maxima or minima. The second derivative test can then be used to determine whether a critical point corresponds to a maximum or a minimum.
By expressing the area as a function of width, we have transformed the problem into a standard optimization problem that can be solved using a variety of mathematical techniques. This step is essential because it allows us to focus on finding the maximum value of a single-variable function, which is a well-understood problem in calculus and algebra. The next step is to use one of the methods described above to find the width that maximizes the area.
Finding the Maximum Area Using Calculus
To find the greatest rectangular area the farmer can enclose, we can use calculus to maximize the area function A(w) = 50w - w^2. This involves finding the critical points of the function by taking its derivative and setting it equal to zero.
The derivative of A(w) with respect to w is:
To find the critical points, we set the derivative equal to zero:
Solving for w:
So, the critical point occurs at w = 25 meters. To confirm that this critical point corresponds to a maximum, we can use the second derivative test. The second derivative of A(w) is:
Since A''(w) = -2 is negative for all values of w, the function A(w) is concave down, and the critical point w = 25 corresponds to a maximum. This confirms that the width of 25 meters will maximize the area of the pen.
Now that we have the optimal width, we can find the corresponding length using the equation l = 50 - w:
Thus, the length is also 25 meters. This means that the rectangle that maximizes the area is actually a square with sides of 25 meters each. This is a general principle: for a given perimeter, the rectangle with the largest area is always a square.
Finally, we can calculate the maximum area by multiplying the optimal width and length:
Therefore, the greatest rectangular area that the farmer can enclose with 100 meters of fencing is 625 square meters.
This calculus-based approach provides a rigorous way to find the maximum area. By finding the critical points of the area function and using the second derivative test, we can confidently determine the dimensions that maximize the enclosed area. The result, a square with sides of 25 meters, highlights the efficiency of a square shape in maximizing area for a given perimeter. This principle has applications in various fields, from designing enclosures to optimizing the layout of rooms and buildings.
The Greatest Rectangular Area
In conclusion, by setting up the perimeter equation, expressing the area as a function of the width, and using calculus to find the maximum value, we have determined the greatest rectangular area that the farmer can enclose with 100 meters of fencing. The optimal dimensions for the pen are a width of 25 meters and a length of 25 meters, resulting in a square shape. The maximum area is:
Therefore, the greatest rectangular area that the farmer can enclose with 100 m of fencing is 625 square meters.
This problem demonstrates a classic optimization scenario, where the goal is to maximize a quantity (in this case, area) subject to a constraint (the fixed amount of fencing). The solution highlights the importance of understanding mathematical relationships and using appropriate techniques to solve real-world problems. The fact that the optimal shape is a square is a well-known principle in geometry and has practical implications in various applications.
The process of solving this problem involved several key steps. First, we translated the problem into mathematical terms by setting up the perimeter equation and expressing the length in terms of the width. This allowed us to write the area as a function of a single variable, the width. Next, we used calculus to find the critical points of the area function, which represent potential maxima or minima. The second derivative test confirmed that the critical point we found corresponds to a maximum. Finally, we calculated the maximum area using the optimal dimensions. This systematic approach is applicable to a wide range of optimization problems.
The result of this problem, 625 square meters, provides the farmer with a concrete answer to their question. By constructing a square pen with sides of 25 meters, the farmer can enclose the largest possible area with the available fencing. This result is not only mathematically sound but also practically useful for the farmer in planning their land use. The combination of mathematical reasoning and practical application makes this problem a valuable example of how mathematics can be used to solve real-world challenges.
Therefore, the final answer is 625 .
