Peter has 1200 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area?

So what exactly would the answer be? im confused..

1200/4 = 300 yards on each side.

A = 300 * 300

A = ?

Finding the Absolute Area....

Area of the fence:
A=x*y

Perimeter of the fence
2x+2y=1200

Then, you solve for one variable
2x=1200-2y
x=600-y <-- plug this back into the Area function and you will get

A=(600-y)y
A=600y-y^2

And I believe you can set that equal to zero
-y^2+600y=0

Then you use the quadratic formula
b=600
a=-1
so y= -600/(2)*(-1)
y=300

That'll give you one side, then plug it back to perimeter equation to get x. Hope this is right!

You're welcome.

To solve this problem, we need to use the concept of optimization. The first step is to identify the variables involved and set up an equation that represents the problem. In this case, we need to find the dimensions of the rectangle that maximize the enclosed area. Let's call the length of the rectangle L and the width W.

The perimeter of a rectangle is given by the formula P = 2L + 2W. In this problem, we are given that there are 1200 yards of fencing, so we can set up the equation:

2L + 2W = 1200

Now, we need another equation that represents the area of the rectangle. The area of a rectangle is given by the formula A = LW. We want to maximize the enclosed area, so we need to find the maximum value of A.

Now, let's solve for L in terms of W using the first equation:

2L = 1200 - 2W

L = 600 - W

Substitute this expression for L in the formula for the area:

A = (600 - W)W

Expand and simplify:

A = 600W - W^2

Now, we have an equation for the area in terms of W. To find the maximum value of A, we need to find the value of W that makes the derivative of A equal to zero. Let's take the derivative of A with respect to W:

dA/dW = 600 - 2W

Set this equal to zero and solve for W:

600 - 2W = 0

2W = 600

W = 300

Now that we have the value of W, we can substitute it back into the equation for L to find the length of the rectangle:

L = 600 - W

L = 600 - 300

L = 300

So, the dimensions of the rectangle that maximize the enclosed area are a length of 300 yards and a width of 300 yards.

Oh i see now! Ok thank you!

It's a square.