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"#