The open box will have the following dimensions: l=6-2x,w=6-2x,h=x

(You take a segment of length x from both corners on a side.)

The volume is lxwxh so:

`V=x(6-2x)(6-2x)`

`=4x^3-24x^2+36x`

To find the maximal volume we take the first derivative and find the critical points by setting equal to zero:

`(dV)/(dx)=12x^2-48x+36`

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The open box will have the following dimensions: l=6-2x,w=6-2x,h=x

(You take a segment of length x from both corners on a side.)

The volume is lxwxh so:

`V=x(6-2x)(6-2x)`

`=4x^3-24x^2+36x`

To find the maximal volume we take the first derivative and find the critical points by setting equal to zero:

`(dV)/(dx)=12x^2-48x+36`

`=12(x-3)(x-1)`

The critical points are x=3 or x=1. From the problem, x=3 will not work as you will have no material left. Therefore x=1 is the only possible solution.

**The volume will be V=(4)(4)(1)=16cu units.**