Finding the Corresponding Length

Oct 13, 2025 by thesweetdesigns.com 125 views

Introduction

When it comes to maximizing the area of a rectangular plot, farmers often face a challenge: how to make the most of the available fencing while ensuring that the plot is as large as possible. In this scenario, we have a farmer with 4000 meters of fencing who wants to enclose a rectangular plot that borders on a river. The farmer does not need to fence the side along the river, which means that the length of the plot will be determined by the river itself. Our goal is to find the largest area that can be enclosed by the farmer using the available fencing.

The Problem

Let's denote the length of the plot as L and the width as W. Since the farmer does not need to fence the side along the river, we only need to consider the three sides of the plot that require fencing. The total length of fencing available is 4000 meters, which means that the sum of the lengths of the three sides is equal to 4000 meters. Mathematically, this can be expressed as:

2W + L = 4000

Maximizing the Area

To maximize the area of the plot, we need to find the optimal values of L and W. The area of the plot is given by the formula:

A = L × W

We want to maximize this area subject to the constraint that 2W + L = 4000. To do this, we can use the method of substitution to express L in terms of W. Rearranging the constraint equation, we get:

L = 4000 - 2W

Substituting this expression for L into the area formula, we get:

A = (4000 - 2W) × W

Expanding and simplifying the expression, we get:

A = 4000W - 2W^2

Finding the Maximum Area

To find the maximum area, we need to find the critical points of the function A = 4000W - 2W^2. This involves taking the derivative of the function with respect to W and setting it equal to zero. The derivative of A with respect to W is:

dA/dW = 4000 - 4W

Setting this derivative equal to zero, we get:

4000 - 4W = 0

Solving for W, we get:

W = 1000

Finding the Corresponding Length

Now that we have found the value of W, we can find the corresponding value of L by substituting W into the expression L = 4000 - 2W. This gives us:

L = 4000 - 2(1000) L = 2000

Calculating the Maximum Area

Now that we have found the values of L and W, we can calculate the maximum area by substituting these values into the area formula A = L × W. This gives us:

A = 2000 × 1000 A = 2,000,000 $m^2$

Conclusion

In this scenario, the farmer can enclose a rectangular plot with a maximum area of 2,000,000 $m^2$ . This is the largest area that can be enclosed by the farmer using the available fencing.

Discussion

The problem of maximizing the area of a rectangular plot is a classic example of an optimization problem. The goal is to find the optimal values of the variables (in this case, L and W) that maximize the area subject to a constraint (in this case, the total length of fencing available). The method of substitution is a useful technique for solving optimization problems like this one.

Final Answer

The final answer is: 2,000,000 $m^2$

Introduction

In our previous article, we explored the problem of a farmer with 4000 meters of fencing who wants to enclose a rectangular plot that borders on a river. We found that the maximum area that can be enclosed is 2,000,000 $m^2$ . In this article, we will answer some of the most frequently asked questions related to this problem.

Q: What is the main constraint in this problem?

A: The main constraint in this problem is that the farmer has 4000 meters of fencing available, and the farmer does not need to fence the side along the river.

Q: How do we maximize the area of the plot?

A: To maximize the area of the plot, we need to find the optimal values of L and W. We can use the method of substitution to express L in terms of W, and then find the critical points of the function A = 4000W - 2W^2.

Q: What is the relationship between the length and width of the plot?

A: The relationship between the length and width of the plot is given by the constraint equation 2W + L = 4000. We can rearrange this equation to express L in terms of W as L = 4000 - 2W.

Q: How do we find the maximum area?

A: To find the maximum area, we need to find the critical points of the function A = 4000W - 2W^2. We can do this by taking the derivative of the function with respect to W and setting it equal to zero.

Q: What is the value of W that maximizes the area?

A: The value of W that maximizes the area is W = 1000.

Q: What is the corresponding value of L?

A: The corresponding value of L is L = 2000.

Q: What is the maximum area that can be enclosed?

A: The maximum area that can be enclosed is 2,000,000 $m^2$ .

Q: Why is this problem an example of an optimization problem?

A: This problem is an example of an optimization problem because we want to find the optimal values of L and W that maximize the area subject to a constraint (the total length of fencing available).

Q: What is the method of substitution?

A: The method of substitution is a technique used to solve optimization problems by expressing one variable in terms of another variable.

Q: What is the derivative of the function A = 4000W - 2W^2?

A: The derivative of the function A = 4000W - 2W^2 is dA/dW = 4000 - 4W.

Q: How do we find the critical points of the function A = 4000W - 2W^2?

A: We can find the critical points of the function A = 4000W - 2W^2 by taking the derivative of the function with respect to W and setting it equal to zero.

Q: What is the significance of the critical points in this problem?

A: The critical points in this problem represent the values of W that maximize the area.

Q: How do we calculate the maximum area?

A: We can calculate the maximum area by substituting the values of L and W into the area formula A = L × W.

Q: What is the final answer to this problem?

A: The final answer to this problem is 2,000,000 $m^2$ .

Conclusion

In this article, we have answered some of the most frequently asked questions related to the problem of a farmer with 4000 meters of fencing who wants to enclose a rectangular plot that borders on a river. We have found that the maximum area that can be enclosed is 2,000,000 $m^2$ .