so hmm if you check the picture below, the area of the base is xx and the top's is xx and the sides, are 4 rectangles, each with area of h*x
thus \(\bf \textit{volume of a rectangular prism}\\\\
V=lwh\qquad
\begin{cases}
l=length\\
w=width\\
h=height\\
----------\\
l=w=x\\
base=l\cdot w\\
\qquad x\cdot x\implies x^2\\
V=40
\end{cases}\implies
\begin{array}{llll}
40=x^2h\\\\
\cfrac{40}{x^2}=\boxed{h}
\end{array}\\\\
-----------------------------\)
\(\bf \begin{array}{llll}
\textit{area of the base}\\\\
A_b=x^2
\end{array}
\qquad
\begin{array}{llll}
\textit{area of the top}\\\\
A_b=x^2
\end{array}
\\\\\\
\textit{area of the sides, or lateral area}\\\\
A_l=xh+xh+xh+xh\implies 4xh\implies 4x\cdot \boxed{\cfrac{40}{x^2}}\implies \cfrac{160}{x}\)
now, the cost function, or C(x), well, we know how much area is going to be used in x-term, so let's just apply the cost to each
\(\bf C(x)=0.37A_b+0.13A_t+0.05A_s\to C(x)=0.37x^2+0.13x^2+\cfrac{8}{x}
\\\\\\
C(x)=0.5x^2+8x^{-1}\)
so.... now just get the derivative of C(x) and zero it out out and check the critcal points for any minima
