If 800 feet of fencing is available, find the maximum area that can be enclosed.

[math_celebrity]

If 800 feet of fencing is available, find the maximum area that can be enclosed.

Perimeter of a rectangle is:
2l + 2w = P

However, we're given one side (length) is bordered by the river and the fence length is 800, so we have:
So we have l + 2w = 800

Rearranging in terms of l, we have:
l = 800 - 2w

The Area of a rectangle is:
A = lw

Plug in the value for l in the perimeter into this:
A = (800 - 2w)w
A = 800w - 2w^2

Take the first derivative:
A' = 800 - 4w

Now set this equal to 0 for maximum points:
4w = 800

Typing this equation into the search engine, we get:
w = 200

Now plug this into our perimeter equation:
l = 800 - 2(200)
l = 800 - 400
l = 400

The maximum area to be enclosed is;
A = lw
A = 400(200)
A = 80,000 square feet