Let s be the side length of the bottom square of the box, and h be its height. Then the volume of the box is
s^2h = 931
=> h = 931/s^2
The areas of the bottom, top, and sides rectangular, and thus can be calculated as the product of their side lengths:
A_"bot" = A_"top" = s^2
A_"side" = sh = s(931/s^2) = 931/s
Now we can calculate the total cost C (in cents) as a function of s:
C = 15A_"bot" + 10A_"top" + 2.5(4A_"side")
=15s^2+10s^2+10(931/s)
=25s^2+9310/s
Our goal is to minimize C. Note that we must have the bounds 0 < s < sqrt(931) for our initial equation to hold. To find the extrema of C(s) = 25s^2+9310/s, we will find where its derivative is 0 on that interval:
C'(s) = 2(25s)+9310(-1/s^2) = 50s-9310/s^2 = 0
=> 50s^3 - 9310 = 0
=> s^3 = 931/5
=>s = (931/5)^(1/3)
So, the only possible minimum could be at s=(931/5)^(1/3). There is also the chance that the cost could decrease as we approach an endpoint, but it's easy to see that as 9310/s->oo as s->0^+, meaning the cost also increases without bound. We can check that s=sqrt(931) results in a higher cost directly:
C((931/5)^(1/3)) = 25(931/5)^(2/3)+9310/(931/5)^(1/3) ~~1773
C(sqrt(931)) = 25(931)+9310/sqrt(931) >= 25(931) > 1773
Thus, the cost is minimized with the dimensions:
"length" = "width" = (931/5)^(1/3)~~5.71" ft"
"height" = 9310/(931/5)^(2/3) ~~ 28.55" ft"
