Fencing perimeters: why is the intuition wrong?

This is from a practice GRE problem.

"A total of x feet of fencing is to form 3 sides of a level rectangular yard. What is the maximum area in terms of x?"

I can do the calculations and take the derivative to see that the area is maximized when the length is twice the width, thus giving $x^2/8$ as the answer. However, I do not see intuitively/geometrically why this should be true. The answer that makes the most sense to me is a square, i.e. $x^2/9$, since it at least "looks" like can start with length zero, and slowly increase until length equals width, increasing area all the time. For fencing surrounding four sides, the answer IS a square, and I don't understand the essence of the difference between these two different cases. Can anyone help me understand this problem better?

• Here's what I find intuitive. Imagine that you have to pay for the length of three of the sides, but the fourth one is free. Then you can afford to make it longer than you would have otherwise, when you had to pay for all four sides. – Rahul Jul 18 '11 at 5:05

If the side of the rectangle not covered by fencing were a mirror, then the fencing together with its reflection should solve the problem of maximizing the area which a rectangle with a perimeter of $2x$ feet can enclose.
If $W$ and $L$ are the width and length of the new rectangle, then $W+L=x$, so $W$ and $L$ can be written as $W=\frac{x}{2}-t$ and $L=\frac{x}{2}+t$ for some $t$ between $0$ and $\frac{x}{2}$. Then $WL=\frac{x^2}{4}-t^2$ is maximized when $t=0$.
• The reflection is a great idea. After that, one can also say the new rectangle must have maximum area subject to its perimeter being $2x$, which means it's a square, which gives that the length of the original rectangle is twice its width, etc. – ShreevatsaR Jul 18 '11 at 6:06