Calculus Problem Involving Area of Rectangle

In my final exam review for calculus there is a problem that asks "Find the area of the largest rectangle having an area of 200 feet. I worked out what I thought was the correct way to do it and got an answer that when plugged into a variety of formulas works perfectly. Here's the problem, it's not what was in the answer guide.

Here's what I did:
Area= Length x Width. P=2L + 2W. P=200
2L+2W=200 ---> 2L=200-2W ---> L=(200-2W)/(2)= 100-W
Area= W(100-W) ---> 100W - W^2 ---> W^2 -100= 0 ---> W^2=100 --> W=10
Length= 100-10=90
Area= 90 x 10= 900 ft

I thought that was the right answer, but the guide says that the correct answer is in fact 2500 Sq. Units. What am I doing wrong?

I assume you mean "area of the largest rectangle with perimeter of 200 feet"
P=2L+2W=200
L+W=100
L=100-W

A=L*W

Now plug in L=100-W into the area equation

A=100W-W^2

Now take the derivative with respect to width- so you get:
dA/dW=100-2W

Set the new equation equal to zero:
100=2W
W=50

Plug in 50 into L+W=100 and you get:
L=50

Remember- because the rate of change of the area with respect to the width equals zero, the area must either be at a maximum or minimum (in this case a maximum)

I'm assuming that you meant a PERIMETER of 200 feet. let P be the perimeter and A be the area:

P = 2l + 2w
200 = 2l + 2w
w = 100 - l

A = lw
A(l) = (l)(100 - l)
A(l) = 100l - l^2

Take the derivative, set it equal to zero:

A ' (l) = 100 - 2l
0 = 100 - 2l
2l = 100
l = 50 feet

w = 100 - l
w = 100 - 50
w = 50 feet

A = lw
A = (50)(50)
A = 2500 square feet

Done!