You should use the notations such that: x for the lengths of square base and y for the height of box, hence, evaluating the volume of the box yields:

`V = x^2*y => 160 = x^2*y => y = 160/x^2`

You need to evaluate the total surface area of box such that:

`A(x) = x^2 + 4x*y`

You should substitute `160/x^2`  for y in equation of `A(x)`  such that:

`A(x) = x^2 + 4x*160/x^2 => A(x) = x^2 + 640/x`

`A(x) = (x^3+160)/x`

You need to optimize the dimensions of the box, hence, you need to find the solutions to the equation `A'(x)=0`  such that:

`A'(x) = ((x^3+160)/x)' =>A'(x) = (3x^2*x - x^3 - 160)/x^2`

`A'(x) = (2x^3 - 160)/x^2`

`A'(x)=0 <=> 2x^3 - 160 = 0 => x^3 - 80 = 0`

Using the formula `(a-b)(a^2+ab+b^2)=a^3-b^3`  yields:

`x^3 - 80 = (x - root(3)80)(x^2 + xroot(3)80 + root(3)(80^2))`

`x^3 - 80 = 0 => x - root(3)80 = 0 => x = root(3)80`

`x^2 + xroot(3)80 + root(3)(80^2) != 0 for x in R`

Hence the base `x = root(3)80 ` and the height `y = 160/(root(3)(80^2))`

Hence, evaluating the dimensions that minimize the cost of the box yields `x = root(3)80 feet ` and `y = 160/(root(3)(80^2)) feet.`