## Calculus: Early Transcendentals 8th Edition

Area = $10A-A^{2}$, $0\lt A\lt10$
Let the lengths of the sides be $A$ and $B$, where $A\neq B$ Perimeter = $2A+2B=20$ $A+B=10, B=10-A$ Area = $AB=A(10-A)=10A-A^{2}$ $\because$ Lengths are positive $A\gt0,$ $B=10-A\gt0, A\lt10$ $\therefore 0\lt A\lt10$