An open-top box uses a minimum of material when the shape of it is half a cube. Two such boxes would form a cube with area 2400 cm², so side area of 2400/6 = 400 cm², and edge dimensions of √(400 cm²) = 20 cm.
The volume of the half-cube is then (1/2)·(20 cm)³ = 4000 cm³.
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Let x represent the edge-length of the square base of the box. Then the height of the side is v/x², and the total area of the open-top box is
... 1200 = x² +4x(v/(x²))
Multiplying by x and subtracting the left side of the equation gives
... x³ -1200x +4v = 0
Differentiating, we have
... 3x² -1200 +4v' = 0
At the maximum volume, v' = 0, so this becomes
... 3x² -1200 = 0
... x² = 400
... x = 20 . . . . . . . . as argued above.
