Rectangle of maximum area with given perimeter

Text

A wire $40$ cm long is bent into a rectangle. Which dimensions maximize the area, and what is that maximum area?

Solution

Half-perimeter $20$ cm: if one side is $x$, the other is $20-x$, with $x\in(0,20)$. Area $A(x)=x(20-x)=20x-x^{2}.$ $A'(x)=20-2x=0\Rightarrow x=10$; since $A''=-2<0$ it is a maximum. The rectangle is the $10\times10$ square, with $A=100\ \mathrm{cm}^{2}$. $\boxed{x=10\ \mathrm{cm}\ (\text{square}),\qquad A_{max}=100\ \mathrm{cm}^{2}}$