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December 17, 2014

December 17, 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)

- Calculus -
**Trisha**, Saturday, September 6, 2014 at 11:33pmOkay, so im not understanding how you went from V'(x)=(12-2x)²+2x(12-2x)(-2) to V'(x)=12(x²-8)... ive done this problem many times and am stuck on this part. i always end up with V'(x)= 12x²-96x+144 => 12(x²-8x+12)

so i end up with x=2 and x=6... which doesn't make sense because x=6 would make the base 0??? Help please!!!

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