Is there a formula or rule that proves that squares are the most

>>8865077

Circles > squares.

learn to differentiate senpai

>>8865086
Forgot to mention rectangle vs square, sorry.

>>8865077
Call the side lengths of the rectangle x and y and the perimeter p. Then you get the optimization problem

[eqn]\begin{align}
\max && xy \\
\text{s.t.} && 2x + 2y = p
\end{align} [/eqn]

Using the method of Lagrange mutliplies, this gives the system:

[eqn] y = \lambda \cdot 2 \\
x = \lambda \cdot 2 \\
2x + 2y = p [/eqn]

With the solution:
[eqn] x = \frac{p}{4} \\
y = \frac{p}{4} \\
\lambda = \frac{p}{8} [/eqn]

Which shows it's indeed a square.

Look up Calculus 1 square optimization

>>8865077

Let perimeter = 4x

all rectangles constructed with this perimeter will have length x+a and width x-a, where a can be any variation in the side length less than x


the area will therefore be (x+a)(x-a)
which is equal to
x^2 - a^2
clearly,since a^2 is always positive, x^2 - a^2 is maximized when a = 0

Yes

To get x * y (area) to be the biggest for a given x + x (or 2x + 2y), x should equal y.

>>8865144
Obviously, that should have said "for a given x + y".

>>8865077
Well, the first thing you need to understand is the geometry is a spectrum.

>>8865077
Difference of 2 squares is a very basic intuitive way to grasp it (comparing a square to a rectangle)

A square's area is essentially a^2 - 0^2, a rectangle's is going to be a^2 - b^2 within the same perimeter, thus a square is always going to have a larger area