## Calculus: Early Transcendentals 8th Edition

Perimeter = $2A+\frac{32}{A}$, $A\gt0$
Let the lengths of the rectangle be $A$ and $B$, where $A\ne B$. Area = $AB=16$ $B=\frac{16}{A}$ Perimeter = $2A+2B=2A+\frac{32}{A}$ $\because$ lengths are positive, $A\gt0$ $B=\frac{16}{A}\gt0$