= Solution
The base costs $ab$, the glass front costs $2ac$, the wooden rear costs $ac$, and the other two walls cost $2bc$. Thus minimize $C=ab+3ac+2bc$ subject to $abc=V>0$. With the <Lagrange multiplier> convention $\nabla C=\lambda\nabla(abc)$, the equations are
$$
b+3c=\lambda bc,\qquad a+2c=\lambda ac,\qquad 3a+2b=\lambda ab.
$$
Multiply these by $a,b,c$, respectively. Comparing their equal right sides gives $ab=3ac=2bc$. Since all dimensions are positive, $a=2c$ and $b=3c$. The volume condition then yields
$$
\boxed{(a,b,c)=(2,3,1)\left(\frac V6\right)^{1/3}.}
$$
This stationary shape is the unique <global minimum>, not merely a candidate: the <arithmetic-geometric mean inequality> applied to the three cost terms gives
$$
C\geq3[(ab)(3ac)(2bc)]^{1/3}=3(6V^2)^{1/3},
$$
with equality exactly at those ratios. Therefore the least cost, in pounds in the specified units, is
$$
\boxed{C_{\min}=3\,6^{1/3}V^{2/3}=18\left(\frac V6\right)^{2/3}.}
$$
This is a <weighted open-box minimization>, with each pair of opposite walls combined into its appropriate positive face weight.
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