Question

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?

Answer 1

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.