Let the side of the square base be x, and height of the dish be h.
Then Volume, V = x^2 * h = 2048 cm^3
h = 2048/x^2 -----------------------------(1)
Also, Surface area of the box, A = x^2 + 4xh ------------------(2)
On putting the value of (1) in (2) we get:
A = x^2 + 4x * 2048/x^2
or A = x^2 + 8192/x
We want to minimize A so,
A' = 2x - 8192/x^2 = 0
or A' = (2x^3 - 8192)/x^2
To construct a box, x cannot be equal to 0 or negative, so:
2x^3 - 8192 = 0
or x^3 = 8192/2 = 4096
or x = 16 (Critical Number)
To verify that x= 16 minimizes A, we use 2nd derivative test.
A'' = 2 + 16384/x^3, which is greater than 0 at x=40, so A is minimum at x=40
h= 2048/x^2
h = 2048/(16^2) =8
Therefore, the least area should be
A = x^2 + 4xh = 16^2 + 4*16*8 = 768 cm^2