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April 19, 2014

April 19, 2014

Posted by **Candy** on Thursday, March 8, 2012 at 4:10pm.

- Calculus -
**MathMate**, Thursday, March 8, 2012 at 7:05pmLet the length of the (four) squares cut from the corner be x.

The box will then have dimensions 12-2x, 12-2x and x once the open box is made.

The volume is therefore:

V(x)=x(12-2x)²

Differentiate with respect to x and equate to zero to get the greatest voume:

V'(x)=(12-2x)²+2x(12-2x)(-2)

=12(x²-8)

Equate to zero and solve for x:

12(x²-8)=0

=>

x=sqrt(8)

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