The optimum shape of such a box is half a cube. The corresponding cube will have a volume of 2×256 ft³ = 512 ft³ = (8 ft)³. Such a box has a square base that is 8 ft on a side. If the height is half that of the cube, it will be 4 ft.
The dimensions of your box will be 8 ft square by 4 ft high.
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If the base dimension is x ft, the area (quantity of material) is
... a = x² + 4x(256/x²)
... a = x² + 1024x⁻¹
Then the derivative of area with respect to x is
... a' = 2x -1024x⁻²
Setting this derivative to zero and solving for x gives the value of x for minimum area.
... 0 = 2x -1024/x²
... 512 = x³
... x = 8 . . . . . . . . same as above.