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