Corinda has 400 feet of square fencing to make a play area. She wants the fenced area to be rectangular.What dimensions should she use in order to enclose the maximum possible area?

rectangle with maximum area is a square. So, the yard is 100x100.

What is "square fencing"?

2 ft

To find the dimensions that will enclose the maximum possible area, we can use the concept of calculus and basic geometry.

Let's assume the width of the rectangular play area is 'w' feet, and the length is 'l' feet.
To maximize the area, we want to make the rectangle as close to a square as possible.

Given that the perimeter of the rectangle is 400 feet, we can write the equation for the perimeter as:
2w + 2l = 400

We need to express one variable in terms of the other, so let's rearrange the equation:
2w = 400 - 2l
w = 200 - l/2

The area of the rectangular play area is given by:
A = w * l

Substituting the expression for w from above, we have:
A = (200 - l/2) * l
A = 200l - (l^2)/2

To find the maximum possible area, we need to find the value of l that maximizes A. We can do this by taking the derivative of A with respect to l, setting it equal to zero, and solving for l.

dA/dl = 200 - l
Setting this equal to zero and solving for l, we get:
200 - l = 0
l = 200

With l = 200, we can substitute this value back into the expression for w:
w = 200 - l/2
w = 200 - 200/2
w = 200 - 100
w = 100

Therefore, to enclose the maximum possible area with 400 feet of square fencing, Corinda should use dimensions of 100 feet by 200 feet.

To find the dimensions that would enclose the maximum possible area within a rectangular fenced area, we can use the concept of calculus.

Let's assume the length of the rectangular fenced area is L and the width is W. According to the problem, we know that the sum of all four sides of the fencing is 400 feet, which can be expressed as:

2L + 2W = 400

We can simplify this equation to:

L + W = 200

To find the maximum possible area, we need to express the area in terms of a single variable, which is L or W. In this case, let's solve for L in terms of W and substitute it into the area formula.

L = 200 - W

Now, the area of the rectangle can be calculated as:

A = L * W

Substituting the value of L from above:

A = (200 - W) * W

Expanding it:

A = 200W - W^2

Now, to find the maximum area, we have to find the value of W that maximizes this formula. This can be done by finding the derivative of A with respect to W and setting it to zero:

dA/dW = 200 - 2W

Setting it to zero:

200 - 2W = 0

Solving for W:

2W = 200

W = 100

Now we can substitute this value of W back into the equation for L to find the length:

L = 200 - W
L = 200 - 100
L = 100

Therefore, the dimensions that Corinda should use to enclose the maximum possible area with the given 400 feet of square fencing is 100 feet by 100 feet (L = 100 ft, W = 100 ft).