A rectangular play area will be enclosed using 100 yards of fencing. What is the maximum area that can be enclosed?
The largest rectangle will be a square, so
4x = 100
x = 25
Area = 25^2 = 625 square yards.
To find the maximum area that can be enclosed, we need to determine the dimensions of the rectangular play area.
Let's assume the length of the rectangular play area is "L" and the width is "W".
The perimeter of a rectangle is given by the formula: Perimeter = 2L + 2W.
Given that the perimeter is 100 yards, we can write the equation: 2L + 2W = 100.
We can simplify the equation by dividing both sides by 2: L + W = 50.
To find the maximum area, we need to maximize the product of the length and width. So we can rewrite the equation as: L = 50 - W.
Now, we can express the area of the rectangle (A) in terms of one variable (width W): A = L * W.
Substituting the value of L from the above equation into the area equation, we have: A = (50 - W) * W.
Expanding this equation, we get: A = 50W - W^2.
To find the maximum value of A, we need to find the vertex of the quadratic equation, which is given by the formula: x = -b/2a.
In our case, a = -1, and b = 50. Substituting these values into the formula, we get: W = -50 / (2*(-1)).
Simplifying this equation, we have: W = 25.
So, the width of the rectangular play area that maximizes the area is 25 yards.
Substituting this value back into the equation L = 50 - W, we get: L = 50 - 25 = 25.
Therefore, the dimensions of the rectangular play area that maximize the enclosed area are a length of 25 yards and a width of 25 yards.
To find the maximum area, we can substitute these values into the area equation: A = L * W = 25 * 25 = 625 square yards.
Thus, the maximum area that can be enclosed using 100 yards of fencing is 625 square yards.
To find the maximum area that can be enclosed, we need to determine the dimensions of the rectangle. Let's denote the width of the rectangle as 'w' and the length as 'l'.
We know that the sum of all sides of a rectangle is equal to the perimeter, which in this case is 100 yards. The formula to calculate the perimeter of a rectangle is P = 2w + 2l.
Given that P = 100, we can substitute it into the formula and solve for l:
100 = 2w + 2l
Rearranging the equation, we get:
2l = 100 - 2w
l = (100 - 2w) / 2
l = 50 - w
Now, we need to find the maximum area by multiplying the length and width: A = w * l. Substitute the previous equation for 'l':
A = w * (50 - w)
A = 50w - w^2
To find the maximum area, we look for the vertex of the parabolic function A = 50w - w^2. The vertex can be found at the x-coordinate of -b/2a for a quadratic equation Ax^2 + Bx + C = 0.
In this case, a = -1, b = 50, and c = 0:
w-coordinate of the vertex = -b/2a = -50 / (2 * -1) = 25
So, the width of the rectangle that results in the maximum area is 25 yards. To find the length, substitute this value back into the equation l = 50 - w:
l = 50 - 25 = 25 yards
Therefore, the maximum area that can be enclosed is given by:
A = w * l = 25 * 25 = 625 square yards.