The least amount of material will be required when the box is half a cube, so its surface area is half that of a cube with volume 64 in³. Such a cube has sides that are 4 in squares, so has an area of 4²×6 = 96 in². Half that is 48 in².
The least amount of material required for an open square box of volume 32 in³ is 48 in².
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Let x represent the edge length of the square base. Then the height of the box is 32/x², and its surface area is
... A = x² + 4x(32/x²) = x² +128/x
The derivative of this is
... dA/dx = 2x -128/x²
and it will be 0 where the area is a minimum:
... 0 = 2x -128/x²
... 64 = x³ . . . . . multiply by x²/2 and add 64. Note this is the answer we had above.
... x = ∛64 = 4, and the area is
... A = 4² + 128/4 = 16 + 32 = 48 . . . . in²
