Squares and rectangles

Take a rectangle imagine you have a fixed perimeter, but that you can vary its sides. Of those rectangles which one has greatest area?

Make a "mental movie" of the rectangle with varying sides, from a very big horizontal side to a small one, you can also look at the picture below, making more rectangles with your imagination.

Let's take the perimeter to be 20 and the sides to be "a" and "b".

Then a+b+a+b=20 and a+b=10

Think about big sides (close to 10) what happens to the area? Why?

Big sides makes bad rectangles because the other side will be forced to be small (maybe almost zero) giving a small final area.

Small and big "a" and "b" gives bad choices so maybe the "equilibrium" a=b is the best choice.

Since a+b=10 we get a=b= 5

Indeed the area in this case will be, 5*5= 25, some more cases below:

A=5*5= 25
A=6*4= 24
A=7*3= 21
A=8*2= 16
A=9*1= 9

So a=b=5 definitively looks like the best choice. We could also try

A=5.1*4.9 = 24.99

and many others, making it even more compelling. Can we visualize what is happening here?



In this picture ABCD is our square and AMNP is some rectangle with same perimeter as the square, we want to show that AMNP has a smaller area than ABCD. But think for a moment, why is that?

Here AN is smaller then side AC by the same amount that the side AM is greater than AB (or CN=BM), since they have the same perimeter. In relation to the square the rectangle AMNP gained the area BMEP and lost the area MECD. Those two rectangles have one side in common BM=CN, but BE is less than NE. Thus the rectangle lost more area than it gained from the square and so the square has greater area!
(Idea of argument from Rademacher and Toeplitz, great book by the way)

A more algebraical way uses a similar idea, comparing the sides of a and b with the square with sides 5,5 we get: a= 5-k and b= 5+k and a+b= 10 then:

Area= A= a*b = (5-k)*(5+k)= 25 - k ²

That is always smaller then 25! So the best is the square or k=0.

Actually we got a little more than that, that the farthest the rectangle is from the square (bigger k) the smaller the area it has.

Problem: Use the geometrical reasoning to get to the same conclusion.

More on that in this post.