Show that among all the rectangles with the same perimeter a, the one with the biggest surface is the square?

Answer 1

Let x,y be sides of recctangle.
So, 2x+2y=a.
Now we want to maximize xy.
Now x+y=a/2, so
y=(a/2)-x.
So, substituting,
x((a/2)-x) = (((ax)/2)-x^2.
Now differinating and set to 0,
(a/2)-2x=0, or
2x=a/2,
So, 4x=a.
SO, substituing back in
x+y=a/2, so
x+y=(4x)/2=2x.
So, y=2x-x=x.
So, x=y means its a square!

Answer 2

P = 2(l+w), so w = p/2 - l
area = l x w = pl/2 - l^2
max when l = p/4
when l = p/4, w = p/4
so it is a square