Answer
(a) $2x$
(b) $x-4$ by $2x-4$, $x\gt 4$
(c) $4x^2-24x+32$
(d) 8 in by 20 in.
(e) $13.05\ in\lt x \lt 14.22\ in$
Work Step by Step
(a) The length is twice its width, thus we have the length as $2x$
(b) The bottom rectangle should have a dimension of $x-4$ by $2x-4$. The restrictions are: $2x-4\gt$ and $x-4\gt0$ which gives $x\gt 4$
(c) The volume is $V=2(x-4)(2x-4)=4x^2-24x+32$
(d) Let $V=320\ in^3$, we have $4x^2-24x+32=320$ or $4x^2-24x-288=0$ which gives $x=12\ in$ (discard the negative answer), thus the dimensions of the bottom part are 8 in by 20 in.
(e) Let $400\lt V\lt 500$, we have $400\lt 4x^2-24x+32\lt 500$ or $92\lt x^2-6x \lt 117$. Solve the left part $x^2-6x-92\gt 0$ to get $x\gt 13.05$. Solve the right part $x^2-6x-117\lt 0$ to get $x\lt 14.22$. Thus the $13.05\ in\lt x \lt 14.22\ in$
