Answer
${$}191.28$
Work Step by Step
Let $l$, $w$ and $h$ be the container's dimensions. We are given that the volume is $V=10$ and $l=2w$:
$V$ = $lwh$
$10$ = $(2w)(w)h$ = $2w^{2}h$
$h$ = $\frac{5}{w^{2}}$
The cost is
$C(w)=10(2w^{2})+6[2(2wh)+2(hw)]+6(2w^{2})$ = $32w^{2}+32wh$= $32w^{2}+32w\left(\frac{5}{w^{2}}\right)$ = $32w^{2}+\frac{180}{w}$
$C'(w)$ = $64w-\frac{180}{w^{2}}$ = $\frac{64w^{3}-180}{w^{2}}$
$C'(w)$ = $0$
$w$ = $\sqrt[3] {\frac{45}{16}}$ is the critical number
There is an absolute minimum for $C$ when $w$ = $\sqrt[3] {\frac{45}{16}}$ since
$C'(w)$ $\lt$ $0$ for $0$ $\lt$ $w$ $\lt$ $\sqrt[3] {\frac{45}{16}}$
$C'(w)$ $\gt$ $0$ for $w$ $\gt$ $\sqrt[3] {\frac{45}{16}}$
The minimum cost is $C\left(\sqrt[3] {\frac{45}{16}}\right)$ = $32\left(\sqrt[3] {\frac{45}{16}}\right)^{2}+\frac{180}{\sqrt[3] {\frac{45}{16}}}$ $\approx$ ${$}191.28$
