## Calculus: Early Transcendentals 8th Edition

$V(x)=4x^{3}-64x^{2}+240x, 0\lt x\lt6$
Length of box = $20-2x$ Width of box = $12-2x$ Height of box = $x$ Volume = Length$\times$Width$\times$Height =$x(20-2x)(12-2x)$ =$4x(10-x)(6-x)$ =$4x(x^{2}-16x+60)$ =$4x^{3}-64x^{2}+240x$ $\because$ All lengths are positive, Length = $20-2x\gt0, x\lt10$ Width = $12-2x\gt0, x\lt6$ Height = $x\gt0$ $\therefore 0\lt x\lt6$