 Solution

Take positive dimensions measured in metres. The volume is $V=lwh$ and the external surface area constraint is $S=lw+2lh+2wh=12$. There is one base and four walls, with no lid. The Lagrange multiplier equations for $V-\lambda (S-12)$ areThe first equation implies $\lambda >0$. Multiply the first by $l$ and the second by $w$ and subtract; this gives $2\lambda h(l-w)=0$, hence $l=w=s$. The last equation gives $\lambda =s/4$. Substituting into the first yields $sh=(s/4)(s+2h)$ and therefore $s=2h$.

The constraint now reads $s^2+4sh=3s^2=12$. ThusFor a global check, apply the arithmetic-geometric mean inequality to $lw$, $2lh$ and $2wh$, whose sum is $12$. Their product is $4V^2\le 4^3=64$, so no feasible box has $V>4$. Equality holds exactly at the dimensions found above. This is the maximum-volume open rectangular box at fixed surface area.