## College Algebra 7th Edition

$A(x)=10x-x^2$, $0\lt x \lt10$
We are told that the rectangle has perimeter of 20 ft ($p=20$). We know that: $A=l*w$ $P=2l+2w=20$ We solve for $l$: $2l+2w=20$ $2l=20-2w$ $l=10-w$ We substitute: $A=l*w=(10-w)w=10w-w^2$ If we use $x$ for $w$, then: $A(x)=10x-x^2$ (And $0\lt x \lt10$ to keep length and width positive.)