Rectangle of maximum area with perimeter 16

Text

Among all rectangles with perimeter $16$, determine the one of maximum area.

Solution

Let $x$ and $y$ be the sides. From the perimeter $16=2x+2y$ we get $y=8-x$. The area is $\mathcal A(x)=x(8-x)=-x^2+8x,$ a downward parabola, maximum at the vertex: $x=4,\qquad y=4.$ The rectangle of maximum area is therefore a square of side $4$, with area $\boxed{\mathcal A_{\max}=16.}$