Question

two equal rectangular lots are enclosed by fencing the perimeter of a rectangular lot and then putting a fence across its middle. if each lot is to contain 432 square feet, what is the minimum amount of fence (in ft) needed to enclose the lots (include the fence across the middle)?

Answer 1

The minimum amount of fence needed to enclose the two equal rectangular lots, including the fence across the middle, is 104 feet.

Step-by-step explanation:

To find the minimum amount of fence needed to enclose the two equal rectangular lots, we need to calculate the perimeter of each individual lot and then add the fence across the middle. Let's assume the length of each lot is L and the width is W. We are given that the area of each lot is 432 square feet, so we can set up the equation LW = 432. Since the two lots are equal, we can say L = W. Substituting that into the equation, we get L^2 = 432. Solving for L, we find L = sqrt(432) = 20.8 feet. Therefore, the length and width of each lot is 20.8 feet. The perimeter of each lot is 2L + 2W, which is 2(20.8) + 2(20.8) = 83.2 feet. Adding the fence across the middle, we get a total perimeter of 83.2 + 20.8 = 104 feet.

Answer 2

For two rectangular lots each with an area of 432 square feet, the minimum amount of fencing required is approximately 103.9 feet, with the lots arranged as squares to minimize the perimeter.

Step-by-step explanation:

To find the minimum amount of fence needed to enclose two rectangular lots each containing 432 square feet, we must consider the dimensions that will minimize the perimeter. We can denote the length of the lot as L and the width of the lot as W, knowing that the area A is equal to L × W = 432 square feet.

For a rectangle, perimeter P is calculated as P = 2L + 2W. Since we need a fence across the middle, the total amount of fence F will be F = P + W. We need to minimize F considering A = 432.

In order to minimize the perimeter, L and W should be as close as possible to each other, which happens when the rectangle is a square. So, W = √432 = 20.78 feet (approx.) and L = √432 = 20.78 feet (approx.). Thus, P = 2×20.78 + 2×20.78 = 83.12 feet, and F = 83.12 + 20.78 = 103.9 feet (approx.). Therefore, the minimum amount of fencing required is approximately 103.9 feet.