Answer
$3(2 v)^{2 / 3}=$Minimum value
Work Step by Step
\[
2 x z+2 y z+x y=f(x, y, z)
\]
The constraint $g(x, y, z)=-v+x y z=0$
We want to require the least amount for its construction $f$ by using Lagrange multipliers,
\[
\lambda \nabla g=\nabla f
\]
\[
\begin{aligned}
\nabla f &=(y+2 z) \hat{i}+(x+2 z) \hat{j}+(2 x+2 y) \hat{k} \\
\nabla g &=y z \hat{i}+x z \hat{j}+x y \hat{k} \\
\because \quad \nabla f &=\lambda \nabla g \\
(y+2 z) \hat{i}+(x+2 z) \hat{j}+(2 x+2 y) \hat{k}) &=\lambda(y z \hat{i}+z x \hat{j}+x y \hat{k})
\end{aligned}
\]
Which is equivalent to the pair of equations
\[
\begin{aligned}
(y+2 z) &=\lambda(y z) \\
(x+2 z) &=\lambda(z x) \\
(2 x+2 y) &=\lambda(x y) \\
\Rightarrow \frac{(y+2 z)}{y z} &=\frac{(x+2 z)}{z x}=\frac{(2 x+2 y)}{x y}
\end{aligned}
\]
$\lambda=0$ leads to $x=y=z=0,$ which is not a box at all.
Thus, solve for $\quad \lambda=1 / z+2 / x$
\[
\begin{aligned}
&1 / z+2 / y =\lambda\\
&2 / y+2 / x= \lambda \\
&=y=2 z \\
2v= x^{3} & \\
f(x, y, z) &=x(x)+2 x(x / 2)+2 x(x / 2) \\
3 x^{2} =f(x, y, z) &\\
3(2 v)^{2 / 3}=f(x, y, z) &
\end{aligned}
\]
