Let the width of the base be $x$ and the height of the box be $y$. The base is a square so its area is $x^2$. Then the volume of the box is "base area times height", so the volume is $V = x^2 y = 40 ft^3$.
The area of the base is $x^2$, so the cost of the base is $0.31 x^2$.
The area of each side is $xy$, so the cost of each side is $0.05 xy$, so the cost of all four sides is $4 \times 0.05xy = 0.2xy$.
The area of the top is $x^2$, so the cost of the top is $0.19 x^2$.
So the total cost is $C = 0.31 x^2 + 0.19 x^2 + 0.2 xy = 0.5 x^2 + 0.2 xy$.
Since $x^2 y = 40$, we have $y = 40/x^2$, so $C = 0.5 x^2 + 0.2 x\times\frac{40}{x^2} = 0.5 x^2 + \frac{8}{x}$. This we can optimise using derivatives.
Advice for similar problems:
A useful strategy is to give names (ie pronumerals) to all the unknowns in your problem. It is easier to talk about things if they have names. In particular, you can start writing down equations.
When contructing equations, a useful strategy is to organise the information you have on the page so that it is easier to see the connections. (I put each part of the box on a separate line, for example.)
Finally, if you know how to do a problem a certain way (using derivatives of a single variable, for example), then once you have everything written down, you can think about the goal you want to get to and see how to get there. We had to get to an equation for $C$ in terms of only $x$ or $y$, so we had to think of a way to make either the $x$ or the $y$ into something in terms of the other. The only other information we had was the volume, so it seemed like a reasonable thing to try.
For length $l$ cm, width $w$ cm and height $h$ cm, the material costs $c$ in dollars are
$c= 20lw+ 6(2lh+2wh)+ 8lw$ $=28lw + 12h(l+w)$
Note that the set volume of the box means that $h=504/lw$, so
$c=28lw + 6048(l+w)/lw$ $=28lw+6048/l + 6048/w$.
Then the partial derivatives on $l$ and $w$ are
$\frac{\delta c}{\delta l} = 28w-6048/l^2 $ and
$\frac{\delta c}{\delta w} = 28l-6048/w^2 $
Solving for zero derivative we get
$l^2w = 216$ and $w^2l= 216 $
and thus $l=w$ at the minimum and we can solve for the dimensions.
Initial answer
It's relatively obvious that the top and bottom should be square, since we're looking to get minimum perimeter for the area.
So given that $s$ is the side length for the square base, the height $h(s)=504/s^2$ and the cost $c(s) = 20s^2+ 6\cdot 4sh(s)+ 8s^2$ $= 28s^2 + 12096/s$.
Then $\frac{dc}{ds} = 56s-12096/s^2$ and we can solve for the minimum at $\frac{dc}{ds} =0$.
Best Answer
Let $x$ be the size of the base and $y$ be the height of the box. The volume is $v=x^2y$. The cost is $ C=75x^2+200xy+20x^2=95x^2+200xy.$ Therefore we have to maximize $x^2y,$ subject to the constraint $95x^2+200xy=400.$ We solve for y as a function of $x$ to get $$ y=(400-95x^2)/200x$$ Upon simplifying and substituting in the formula for $v$ we find $$v=x^2(400-95x^2)/200x =2x-(19/40)x^3.$$ $v'=0$ implies $x=\sqrt {80/57}.$ substituting in the expression for $y$ we get $y=19\sqrt {57/5}.$ therefore the maximum volume is $$v_{max} =1444\sqrt {57/5}$$