Answer
(a) $640-2x$
(b) $0\lt x \lt 320\ ft$
(c) $-2x^2+640x$
(d) between 57.04 ft and 85.17 ft, or between 234.83 ft and 262.96 ft.
(e) $160\ ft$ by $320\ ft$, $51,200\ ft^2$
Work Step by Step
(a) As the total fencing is $640\ ft$, we have the remaining side as $640-2x$
(b) We need to have $x\gt0$ and $640-2x\gt$, thus $0\lt x \lt 320\ ft$
(c) The area is $A=x(640-2x)=-2x^2+640x$
(d) Let $30000\lt A \lt 40000$, we have $30000\lt -2x^2+640x \lt 40000$. Solve $2x^2-640x+30000\lt 0$ to get $x$ between 57.04 ft and 85.17 ft. $2x^2-640x+40000\gt 0$ to get $x$ between 234.83 ft and 262.96 ft.
(e) Let $a=-2, b=640$, the maximum area happens at $x=-\frac{b}{2a}=-\frac{640}{2(-2)}=160\ ft$, the bottom side is $640-2(160)=320\ ft$, and the maximum area is $A=-2(160)^2+640(160)=51,200\ ft^2$
