# 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?

Apr 27, 2015

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

So it's volume will be
$V \left(h\right) = h \times \left(12 - 2 h\right) \times \left(12 - 2 h\right)$
$= 4 {h}^{3} - 48 {h}^{2} + 144 h \text{ (square inches)}$

$\frac{\mathrm{dV}}{\mathrm{dh}} = 12 {h}^{2} - 96 h + 144$

Critical points occur when the derivative ($\frac{\mathrm{dV}}{\mathrm{dh}}$) is zero.

$12 {h}^{2} - 96 h + 144 = 0$

${h}^{2} - 8 h + 12 = 0$

$\left(h - 2\right) \left(h - 6\right) = 0$

$h = 6$ 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
$6 \times 6$ inches from the corners of the larger sheet.