An open -top box is to be made by cutting small congruent squares from the corners of a 12-by12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?

1 Answer
Apr 27, 2015

If the squares cut from the corners are #hxxh# inches
The open-top box will have
a height of #h#
a width of #12-2h#
and a length of #12-2h#

So it's volume will be
#V(h) = hxx(12-2h)xx(12-2h)#
#= 4h^3-48h^2+144h " (square inches)"#

#(dV)/(dh) = 12h^2-96h+144#

Critical points occur when the derivative (#(dV)/(dh)#) is zero.

#12h^2-96h+144 = 0#

#h^2-8h+12 = 0#

#(h-2)(h-6)=0#

#h=6# would result in widths and lengths of zero (and therefore a volume of zero), so it is obviously not the critical point for the maximum.

Therefore the maximum volume is achieved by by cutting out squares that are
#6xx6# inches from the corners of the larger sheet.