TITLE: Fencing perimeters: why is the intuition wrong?
QUESTION [8 upvotes]: 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?

REPLY [14 votes]: 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 $t=0$.

REPLY [2 votes]: Roughly speaking, these sorts of problems tend to have some kind of balance and symmetry to them.  As you said, when you're fencing all four sides, each side should be the same length.  Here, however, since you're only fencing three sides, to use the same amount of fencing in each of the two directions means that the one side has to be twice as long as the other.