Prove that the area of a rectangle is less than that of the square with the same perimeter.
Answers
My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.