Four squares from the corners of a square plate are removed so that it can be turned into a open cubical box. Find the length of a square to be removed in terms of the length of the main square so that the volume of the box is a maximum.

Let the length of the plate be l and that of a small square be x.

If the volume is V,

V = x(l - 2x)^{2}

= x(l^{2} - 4lx + 4x^{2})

= l^{2}x - 4lx^{2} + 4x^{3}

dV/dx = l^{2} - 8lx + 12x^{2}

At the peak of the graph of V against x, dV/dx = 0

12x^{2} - 8lx + l^{2} = 0

x = 8l ± √ 64l^{2} - 48l^{2} / 24

x = 8l ± √ 16l^{2} / 24

x = 8l ± 4l / 24

x = l/2 or l/6

Since l/2 is not possible, x = l/6

Therefore, to maximize the volume, a square of side l/6 should be cut off from each corner of the plate.

If the length is 12 cm, the model is as follows:

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