Question:
A rectangular garden has a length that is twice its width. The dimensions are increased so that the perimeter is doubled and the new shape is a square with an area of 3600 square feet. What was the area of the original garden, in square feet?

Answer:
Let $w$ be the width of the original rectangular garden.  The perimeter of the rectangle is $2(w+2w)=6w$, so the perimeter of the square is $12w$.  The dimensions of the square are $3w\times 3w$ and its area is $(3w)(3w)=9w^2$, so set $9w^2=3600\text{ ft.}^2$ to find $w^2=400$ square feet.  The area of the original rectangle is $(2w)(w)=2w^2=2\cdot400\text{ ft.}^2=\boxed{800}$ square feet.