= Solution
At a regular constrained extremum of $f$ on $g=0$, the tangent <derivatives> of $f$ vanish. Since $\nabla g$ is normal to the constraint surface, the <Lagrange multiplier> condition is
$$
\nabla f=\lambda\nabla g,
\qquad g=0.
$$
One solves these equations and then compares the resulting candidates, including any boundary or singular cases.
Let the base have dimensions $x$ and $y$, and let the height be $z$. Measure cardboard relative to the thickness of the front and back. The weighted amount used is
$$
A=3xy+2xz+4yz,
$$
because the bottom has triple thickness, the two front and back faces have ordinary thickness, and the two side faces have double thickness. The constraint is $xyz=3$.
The multiplier equations are
$$
3y+2z=\lambda yz,
\qquad
3x+4z=\lambda xz,
\qquad
2x+4y=\lambda xy.
$$
They imply
$$
3xy=2xz=4yz.
$$
Thus $x=2y$ and $z=3y/2$; imposing $xyz=3$ gives $y=1$. Therefore
$$
\boxed{x=2,\qquad y=1,\qquad z=\frac32}.
$$
This is the global minimum: by the <arithmetic-geometric mean inequality>,
$$
A\geq3\sqrt[3]{(3xy)(2xz)(4yz)}
=3\sqrt[3]{24(xyz)^2}=18,
$$
and equality holds at these dimensions. This is an instance of <weighted open-box minimization>.
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