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March 4, 2015

March 4, 2015

Posted by **Anonymous** on Wednesday, April 27, 2011 at 12:35pm.

- Math -
**MathMate**, Wednesday, April 27, 2011 at 12:52pmLet squares of size x" be cut from the corners.

Volume of (open) box

=V(x)

=height*length*width

=x(11-2x)(7-2x)

=4x^3-36x^2+77x

For maximum (or minimum), equate derivative to zero:

dV(x)/dx = 12x²-72x+77=0

Solve for x to get

x=3±(√93)/6

=1.39 or 4.61 (approximately)

4.61 is clearly not a feasible solution (because 2*4.61 > 7") and will be rejected.

So the cut-outs will be squares of 1.39" (approximately).

Now verify that the solution so obtained is a maximum by ensuring that d²V(x)/dx² < 0:

d²V(x)/dx² = 24x-72 = -38.6 <0 OK.

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