Question

How to show that of all the rectangles with a given perimeter, the one with greatest area is a square ?

Answer 1

Use algebra and calculus as follows (note that calculus is actually unnecessary here if you know how to optimize quadratic functions using the method of completing the square).

Explanation:

Let `P` be the fixed perimeter of a rectangle with length `x` and height `y`, so that `P=2x+2y`. The area is `A=xy`. We can write this as a function of `x` by solving `P=2x+2y` for `y` and substituting: `P=2x+2y\Rightarrow 2y=P-2x\Rightarrow y=P/2-x`

`\Rightarrow A=f(x)=x(P/2-x)=(P/2)x-x^2` (for `0 < x < P/2`).

Don't be bothered by the fact that `P` is in this last equation. It's just a constant.

The derivative is `(dA)/dx=f'(x)=P/2-2x`. This is zero when `x=P/4` and, in fact, changes sign from positive to negative as `x` increases through `x=P/4`. By the First Derivative Test, this implies that `x=P/4` gives a local maximum value of the area `A=f(x)`. In fact, since `f` is quadratic, it's actually a global maximum.

When `x=P/4`, then `y=P/2-P/4=P/4` as well. This implies that the dimensions of the rectangle are all equal and it's actually a square. In other words, for all rectangles of a given perimeter `P`, the square of side length `P/4` has the greatest area.