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The dimensions of the cardboard $\"12\"$ in. by $\"20\"$ in.

Observe the figure:

The height of the box is $\"x\"$ in.

Width of the box = width of the cardboard $\"-2x=12-2x\"$ .

Length of the box = length of the cardboard $\"-2x=20-2x\"$ .

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Volume of the box = Length x Width x Height.

Volume of the box $\"=(20-2x)(12-2x)(x)\"$

$\"=(240-40x-24x+4x^2)(x)\"$

$\"=x(240-64x+4x^2)\"$

$\"=240x-64x^2+4x^3\"$

$\"=4x^3-64x^2+240x\"$

The volume of the box as a function of $\"x\"$ is $\"4x^3-64x^2+240x\"$ .

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Domain :

All possible values of $\"x\"$ is the domain of the function.

Volume of the box should be greater than zero.

So $\"L>0\"$ , $\"W>0\"$ and $\"H>0\"$ .

$\"20-2x>0\"$ , $\"12-2x>0\"$ and $\"x>0\"$ .

$\"x<10\"$ , $\"x<6\"$ and $\"x>0\"$ .

The domain of the function is $\"0<x<6\"$ .

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The volume of the box as a function of $\"x\"$ is $\"4x^3-64x^2+240x\"$ , $\"0<x<6\"$ .