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?

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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?

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Answer 1 · 2015-04-27T17:12:06

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

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

$\frac{d V}{d h} = 12 h^{2} - 96 h + 144$ $(dV)/(dh) = 12h^2-96h+144$

Critical points occur when the derivative ( $\frac{d V}{d h}$ $(dV)/(dh)$ ) is zero.

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

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

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

$h = 6$ $h=6$ 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
$6 \times 6$ $6xx6$ inches from the corners of the larger sheet.

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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

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