A rectangular park is 60 yards wide and 100 yards long. Give the length and width of another rectangular park that has the same perimeter but a larger area.
Let's call the width of the second rectangular park x and the length y.
The perimeter of the first park is calculated by adding the lengths of all four sides:
2 * (60 + 100) = 320 yards
To have the same perimeter, the second park must also have a perimeter of 320 yards.
The perimeter of a rectangle is given by the formula P =2 * (length + width).
Therefore, 320 = 2 * (x + y)
Dividing by 2 gives:
x + y = 160
The area of a rectangle is given by the formula A = length * width.
The area of the first park is 60 * 100 = 6000 square yards
To find the dimensions of the second park with a larger area, we want to maximize the product of x and y while keeping x + y = 160.
To do this, we can use the concept of the quadratic function.
Let's denote the area of the second park as A2. This means A2 = x * y.
Since x + y = 160, we can rearrange the equation to get x = 160 - y.
Now, we can express the area in terms of y: A2 = (160-y) * y
Expanding the equation:
A2 = 160y - y^2
We want to find the maximum value of A2.
To do this, we need to find the vertex of the quadratic equation A2 = -y^2 + 160y.
The x-coordinate of the vertex can be found using the formula x = -b/2a. In this case, a = -1 and b = 160.
x = -160 / (2 * -1) = 80
To find the corresponding y-coordinate, we substitute x = 80 into the equation A2 = -y^2 + 160y:
A2 = -(80)^2 + 160(80) = 6400 square yards
Therefore, the length and width of the second rectangular park, which has the same perimeter but a larger area, are 80 yards and 80 yards.