Maximum area at fixed perimeter

The first of the four standard maturità problems: among all rectangles of given perimeter, the one of maximum area is the square.

Example — Maximum area at fixed perimeter

Among all rectangles of fixed perimeter $2p$, which has the maximum area?

Solution

Putting $x$ = base, $y$ = height, constraint $2x+2y=2p \Rightarrow y=p-x$ (with $0<x<p$). Area: $A(x)=x(p-x)=px-x^2, \qquad A'(x)=p-2x=0 \Rightarrow x=\frac{p}{2},$ so $y=p/2$ as well. Since $A''=-2<0$ it is a maximum. Conclusion: the square of side $p/2$ maximises the area; $A_{\max}=\boxed{p^2/4}$.

Links

Topics: Derivatives
Concepts: Objective function · Maximum and minimum
Methods: Maximum minimum via derivative
Skills: Analytic geometry · Modelling