A solid rectangular brick is to be made from 1 cubic feet of clay. The length of brick must be 3 times as long as it is wide. The width of brick for which it will have minimum total surface area is a. Then a³ is

Let width of a rectangular brick is x
  length of a rectangular brick is 3x
Height of a rectangular brick is h
Volume of the rectangular brick  $\left(\mathrm{x}\right)\left(3\mathrm{x}\right)\left(\mathrm{h}\right)=1$
 $\Rightarrow 3{\mathrm{x}}^2\mathrm{h}=1\Rightarrow \mathrm{h}=\frac{1}{3{\mathrm{x}}^2}.................\left(1\right)$
Total surface area of the brick $\mathrm{S}=2\times [\left(\mathrm{x}\right)\left(3\mathrm{x}\right)+\left(3\mathrm{x}\right)\left(\mathrm{h}\right)+\left(\mathrm{h}\right)\left(\mathrm{x}\right)]$
$$\begin{gathered} \Rightarrow \mathrm{S}=6{\mathrm{x}}^2+8\mathrm{hx} \\ \Rightarrow \mathrm{S}\left(\mathrm{x}\right)=6{\mathrm{x}}^2+8\mathrm{x}\left(\frac{{\displaystyle 1}}{{\displaystyle 3{\mathrm{x}}^2}}\right) \\ \Rightarrow \mathrm{S}=6{\mathrm{x}}^2+\frac{8}{3\mathrm{x}} \\ \Rightarrow {\mathrm{S}}^{\text{'}}\left(\mathrm{x}\right)=12\mathrm{x}-\frac{8}{3{\mathrm{x}}^2} \end{gathered}$$
For maxima or minima ${\mathrm{S}}^{\text{'}}\left(\mathrm{x}\right)=0\Rightarrow 36{\mathrm{x}}^2=8\Rightarrow {\mathrm{x}}^3=\frac{2}{9}\Rightarrow \mathrm{x}={\left(\frac{{\displaystyle 2}}{{\displaystyle 9}}\right)}^{\frac{1}{3}}$
$\begin{array}{l}{\mathrm{S}}^{\prime \prime }\left(\mathrm{x}\right)=12+\frac{8}{{\mathrm{x}}^3}\\ {\mathrm{S}}^{\prime \prime }\left({\left(\frac{2}{9}\right)}^{\frac{1}{3}}\right)=12+\frac{8}{\frac{2}{9}}=48>0\end{array}$
$\therefore$surface area is minimum when width $\mathrm{x}={\left(\frac{{\displaystyle 2}}{{\displaystyle 9}}\right)}^{\frac{1}{3}}=\mathrm{a}\Rightarrow {\mathrm{a}}^3=\frac{2}{9}$