Maximizing the Area of a Rectangular Field with Limited Fencing

You have been provided with 3000 feet of fencing to enclose a rectangular field. In this guide, we will explore how to maximize the area of the field using the available fencing, and we will apply mathematical concepts to find the optimal dimensions.

Part a: Expressing the Area as a Function of the Length

Given that the perimeter of the rectangle is 3000 feet, we can express the relationship between the length and width. Let the length of the rectangle be (x) and the width be (w). Since the perimeter is given by the formula (2l 2w 3000), we can simplify this to:

[2x 2w 3000]

Dividing the entire equation by 2:

[x w 1500]

From this, we can express the width (w) in terms of the length (x):

[w 1500 - x]

The area (A) of the rectangle is given by the product of its length and width:

[A xw x(1500 - x) 150 - x^2]

Part b: Finding the Optimal Length for Maximum Area

To find the value of (x) that maximizes the area, we need to take the derivative of the area function with respect to (x) and set it to zero:

[frac{dA}{dx} 1500 - 2x]

Setting the derivative equal to zero to find the critical points:

[1500 - 2x 0]

[2x 1500]

[x 750]

This critical point represents a possible maximum. To ensure that this is indeed a maximum, we take the second derivative of the area function:

[frac{d^2A}{dx^2} -2]

Since the second derivative is negative, the function has a maximum at (x 750). Thus, the length (x) that maximizes the area is 750 feet.

Part c: Calculating the Maximum Area

With the optimal length of 750 feet, we can now calculate the corresponding width and the maximum area:

[w 1500 - x 1500 - 750 750]

Since the length and width are equal, the rectangle is a square, and the maximum area is:

[A 750 times 750 562500 text{ square feet}]

Conclusion

Using the given 3000 feet of fencing, the area of the rectangular field is maximized when it is a square with each side measuring 750 feet, resulting in a maximum area of 562,500 square feet. This is a common outcome because, among all rectangles with a fixed perimeter, a square offers the largest area.

Additional Notes

It is worth noting that the relationship between the length and width of a rectangle with a given perimeter is essential for maximizing its area. For a fixed perimeter, the area is maximized when the rectangle is a square. This is a fundamental concept in geometry and calculus, showcasing the beauty of mathematical optimization problems.