 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 areMultiply 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 yieldsThis stationary shape is the unique global minimum, not merely a candidate: the arithmetic-geometric mean inequality applied to the three cost terms giveswith equality exactly at those ratios. Therefore the least cost, in pounds in the specified units, isThis is a weighted open-box minimization, with each pair of opposite walls combined into its appropriate positive face weight.