$L$ = length, $W$ = width, $H$ = height, $V$ = volume, $C$ = cost
As established in your constraints, $L = 2W$
$V = LWH = 10$
$C = 10LW + 6(2LH + 2WH) = 10LW + 12LH + 12WH$
The above comes from the base being $LW$ with a cost of ten. The walls are $LH$ and $WH$, of which there are 2 each with a cost of 6.
Using the relationship between length and width, you can start substituting variables.
$V = (2W)WH = 2HW^2 = 10$
$C = 10(2W)W + 12(2W)H + 12WH = 20W^2 + 24WH + 12WH = 20W^2 + 36WH$
Now you have a relation for width and height, so you can use that to sub as well.
$2HW^2 = 10$
$H = 10/(2W^2) = 5/W^2$
$C = 20W^2 + 36WH = 20W^2 + 36W(5/W^2) = 20W^2 + 180/W$
From there you find the minimum of the equation where $W > 0$
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.
Best Answer
At any local minimum not on a boundary, the derivative is $0$. There is no boundary in this problem, so finding where the derivative is $0$ gives you all the local minima.
The absolute minimum is also a local minimum. So to find it, you find all the local extrema, evaluate the function at each of them, and choose the smallest value. That have has to be the absolute minimum. In this problem, the derivative is $0$ at only one location: $x = 10$. So assuming this point is a local minimum, it must be the absolute minimum.
So the only thing more that you need to do is confirm that $x = 10$ is indeed a minimum (it could also be a maximum, or an inflection point). The easiest way to do that is to take the second derivative: $$C''(x) = \frac{16000}{x^3} + 8$$ In particular $C''(10) = 24 > 0$. Since it is positive, $x = 10$ is a minimum, and because it is the only minimum, it is the absolute minimum.