find the maximum area of a rectangle whose perimeter is 50 yards
(50/4)^2
To find the maximum area of a rectangle given its perimeter, you can use the concept of maximizing a function.
Let's assume the length of the rectangle is L and the width is W. The perimeter, P, is given by the equation:
P = 2L + 2W
Since the perimeter is fixed at 50 yards, we can write the equation as:
50 = 2L + 2W
Now, we need to express the area of the rectangle in terms of a single variable. The area, A, is given by the equation:
A = L * W
We can rewrite this equation as:
A = L * (50 - 2L) / 2
We can simplify this equation further:
A = (50L - 2LĀ²) / 2
To find the maximum area, we need to take the derivative of the area equation with respect to L and set it equal to zero:
dA/dL = 50/2 - 4L/2 = 0
25 - 2L = 0
Solving this equation, we get:
2L = 25
L = 25/2
L = 12.5
Now, substitute the value of L back into the perimeter equation to find the width:
50 = 2(12.5) + 2W
50 = 25 + 2W
2W = 25
W = 25/2
W = 12.5
Therefore, the dimensions of the rectangle with maximum area are L = 12.5 yards and W = 12.5 yards. The maximum area is:
A = L * W
A = 12.5 * 12.5
A = 156.25 square yards