w = width

h = height

The length is equal to the width

Area of top and bottom = 2w^2

Area of sides = 2wh + 2wh = 4wh

So

Total cost `= 2w^2*2 + (4wh)1 = 4w^2 + 4wh`

The total volume is `30cm^3 = w^2h`

So `h=30/(w^2)` substituting we get

Total cost `= 2w^2 + 4w*30/w^2 = 2w^2 + 120/w`

`(d(Total cos))/(dw) = 4w - 120/w^2`

`(d(Total cost))/(dw) = 0` when

`4w - 120/w^2 = 0`

` 4w = 120/w^2`

`w^3 = 30`

`w = root(3)(30)`

`h = 30/w^2 = 30/(root(3)(30))^2 = root(3)(30)`

I am not sure what your comment about rationals. You can estimate these numbers to any number of decimals. That would be rational answers. The answer is w = 3.107 cm, h = 3.107 cm.

The dimensions of the box are : L, W, and H

We know that L= W.

==> V = L*H*W = L^2 H = 30 ==>

==> H = 30/L^2

==> Surface area for the top and bottom = L*W*2 = 2L^2

==> Cost for top and bottom = 2L^2 * 2 = 4L^2

==> Surface area for the sides = 2HW + 2HL = 2HL+2HL = 4HL

==> Cost for sides = 4HL*1 = 4HL

But H= 30/L^2

==> cost for sides = 4L *30/L^2 = 120/L

==> Total cost = 2L^2 + 120/L

==> f(L) = 2L^2 + 120/L

Now we will find the minimum points:

We will find the derivative;s zeros which given the extreme value.

==> f'(L)= 4L -120/L^2 = 0

==> 4L = 120/L^2

==> 4L^3 = 120

==> L^3 = 30

==> L= 3.12

==> H= 30/30= 1

==> W = L = 3.12

**==> The dimensions are: 3.12, 3.12 , and 1.**