A rectangular lot is 105 yards long and 60 yards wide. Give the length and width of another rectangular lot that has the same perimeter but a larger area
Let's start by finding the perimeter of the first rectangular lot.
Perimeter = 2(length + width)
Perimeter = 2(105 + 60)
Perimeter = 2(165)
Perimeter = 330
Now, let's assume the length of the second rectangular lot is L and the width is W. We want to find the length and width of the second rectangular lot in such a way that the perimeter is 330 yards and the area is larger than the first lot.
For a perimeter of 330 yards, the equation is as follows:
Perimeter = 2(L + W)
330 = 2(L + W)
Dividing both sides by 2, we get:
165 = L + W
Now, we want to find the length and width of the second rectangular lot with a larger area. Since the area of a rectangular lot is given by the product of its length and width, we can express the area of the second lot as:
Area = L * W
To find the dimensions of the second lot with the largest possible area, we need to determine the values of L and W that satisfy the equation 165 = L + W while maximizing the value of the area.
Let's start by considering the case where L = W. In this scenario, L and W are each half of their sum, which means L = W = 165 / 2 = 82.5.
Now, let's check the area when L = 82.5 and W = 82.5:
Area = L * W
Area = 82.5 * 82.5
Area = 6806.25 square yards
So, the length and width of the second rectangular lot, with the same perimeter as the first lot but a larger area, are 82.5 yards and 82.5 yards, respectively.