Finding the Maximum Enclosed Area

Oct 13, 2025 by magicalceylontours.com 15 views

Introduction

When designing a rectangular dog park, one of the primary concerns is maximizing the enclosed area while working within a limited budget for fencing. In this scenario, we have 500 feet of fencing available to create a rectangular dog park split into three sections of the same size. Our goal is to determine the dimensions of the rectangular dog park that will produce the greatest possible enclosed area.

Problem Formulation

Let's denote the length of each section of the dog park as x and the width as y. Since the dog park is split into three sections of the same size, the total length of the dog park will be 3x, and the total width will remain y. The perimeter of the dog park, which is the total length of fencing required, is given by the formula:

P = 2(3x) + 2y

Since we have 500 feet of fencing available, we can set up the equation:

2(3x) + 2y = 500

Simplifying the equation, we get:

6x + 2y = 500

Mathematical Optimization

To maximize the enclosed area of the dog park, we need to find the optimal values of x and y. The enclosed area of the dog park is given by the formula:

A = xy

We can use the method of substitution to solve for x and y. Rearranging the equation 6x + 2y = 500, we get:

y = (500 - 6x) / 2

Substituting this expression for y into the equation for the enclosed area, we get:

A = x((500 - 6x) / 2)

Simplifying the equation, we get:

A = 250x - 3x^2

Finding the Maximum Enclosed Area

To find the maximum enclosed area, we need to find the critical points of the function A = 250x - 3x^2. We can do this by taking the derivative of the function with respect to x and setting it equal to zero:

dA/dx = 250 - 6x = 0

Solving for x, we get:

x = 250 / 6

x = 41.67

Finding the Optimal Dimensions

Now that we have found the optimal value of x, we can substitute it into the equation y = (500 - 6x) / 2 to find the optimal value of y:

y = (500 - 6(41.67)) / 2

y = (500 - 250) / 2

y = 125

Conclusion

In conclusion, the optimal dimensions of the rectangular dog park that will produce the greatest possible enclosed area given 500 feet of fencing are:

Length: 125 feet
Width: 41.67 feet

The maximum enclosed area of the dog park is:

A = xy

A = (125)(41.67)

A = 5233.75

Units

The units of measurement for the dimensions of the dog park are feet, and the units of measurement for the enclosed area are square feet.

References

[1] Calculus: Early Transcendentals, 8th edition, by James Stewart
[2] Optimization Techniques, by S. S. Rao

Additional Information

The optimal dimensions of the dog park are a compromise between the length and width of the dog park. Increasing the length of the dog park will increase the enclosed area, but it will also increase the perimeter of the dog park, which is the total length of fencing required.
The optimal dimensions of the dog park are a function of the available fencing. If more fencing is available, the optimal dimensions of the dog park will change.
The enclosed area of the dog park is a function of the length and width of the dog park. Increasing the length or width of the dog park will increase the enclosed area, but it will also increase the perimeter of the dog park.
Optimizing a Rectangular Dog Park: Maximizing Enclosed Area with Limited Fencing ===========================================================

Q&A: Frequently Asked Questions

Q: What is the optimal shape of a rectangular dog park to maximize the enclosed area? A: The optimal shape of a rectangular dog park to maximize the enclosed area is a square. However, in this scenario, we are working with a rectangular shape split into three sections of the same size.

Q: How do I calculate the maximum enclosed area of a rectangular dog park? A: To calculate the maximum enclosed area of a rectangular dog park, you need to use the formula A = xy, where x is the length and y is the width of the dog park. You also need to use the equation 6x + 2y = 500 to find the optimal values of x and y.

Q: What is the relationship between the length and width of a rectangular dog park and the enclosed area? A: The length and width of a rectangular dog park are inversely related to the enclosed area. Increasing the length or width of the dog park will increase the enclosed area, but it will also increase the perimeter of the dog park.

Q: How do I find the optimal dimensions of a rectangular dog park given a limited amount of fencing? A: To find the optimal dimensions of a rectangular dog park given a limited amount of fencing, you need to use the equation 6x + 2y = 500 to find the optimal values of x and y. You can then substitute these values into the formula A = xy to find the maximum enclosed area.

Q: What are the units of measurement for the dimensions of a rectangular dog park? A: The units of measurement for the dimensions of a rectangular dog park are feet.

Q: What are the units of measurement for the enclosed area of a rectangular dog park? A: The units of measurement for the enclosed area of a rectangular dog park are square feet.

Q: Can I use this method to optimize a dog park with a different shape? A: No, this method is specifically designed for rectangular dog parks. If you want to optimize a dog park with a different shape, you will need to use a different method.

Q: Can I use this method to optimize a dog park with a different amount of fencing? A: Yes, you can use this method to optimize a dog park with a different amount of fencing. Simply substitute the new amount of fencing into the equation 6x + 2y = 500 and solve for x and y.

Q: What are the limitations of this method? A: The limitations of this method are that it assumes a rectangular shape for the dog park and that the amount of fencing is fixed. If you want to optimize a dog park with a different shape or a different amount of fencing, you will need to use a different method.

Conclusion

In conclusion, optimizing a rectangular dog park to maximize the enclosed area with limited fencing requires careful consideration of the length and width of the dog park. By using the equation 6x + 2y = 500 and the formula A = xy, you can find the optimal dimensions of the dog park and the maximum enclosed area.