Final answer:
To find the dimensions of the rectangle that maximize the enclosed area, use the fact that the perimeter of the rectangle is 480 yards. The dimensions of the rectangle that maximize the enclosed area are a length of 120 yards and a width of 120 yards. The maximum area is 14,400 square yards.
Step-by-step explanation:
To find the dimensions of the rectangle that maximize the enclosed area, we can use the fact that the perimeter of the rectangle is 480 yards. Let's denote the length of the rectangle as 'L' and the width as 'W'. We have the equation 2L + 2W = 480, which simplifies to L + W = 240. To maximize the area, we need to maximize the product L * W. Since L + W = 240, we can rewrite this equation as L = 240 - W. Now we can substitute this into the area equation: (240 - W) * W = 240W - W^2. To find the maximum area, we need to find the vertex of the quadratic function. The vertex occurs at W = -b/2a, which in this case is W = -240/-2 = 120. Plugging this value back into the equation, we find that the dimensions of the rectangle that maximize the enclosed area are a length of 120 yards and a width of 120 yards. The maximum area is then 120 * 120 = 14,400 square yards.
To find the dimensions of a rectangle that maximize the enclosed area using a fixed amount of fencing, we use the mathematical principle that for a given perimeter, a square provides the maximum area. With 480 yards of fencing available for all four sides, this means each side of the square would be 480 yards ÷ 4, which is 120 yards. So, a rectangle with dimensions that maximize the area would be a square with sides of 120 yards each. Therefore, the maximum area that can be enclosed is 120 yards × 120 yards which equals 14,400 square yards.