Rettangolo di perimetro 24 con area massima

Text

Among all rectangles with perimeter $24$, find the one of maximum area and the value of that area.

Solution

If one side is $x$, the other is $12-x$ (since the half-sum of the sides is $12$), with $0<x<12$. The area is $A(x)=x(12-x)$. By the arithmetic–geometric mean inequality applied to the two positive factors $x$ and $12-x$: $\sqrt{x(12-x)}\le\frac{x+(12-x)}{2}=6\implies A(x)=x(12-x)\le 36.$ Equality occurs for $x=12-x$, i.e. $x=6$: the rectangle is a square of side $6$. $\boxed{A_{\max}=36\ \text{ for } x=6.}$