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?

Question

Evelyn Agard · Accepted Answer

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.

Emily Ammerman · Answer

undefined To maximize the volume of the box, we need to maximize the area of the base while also maximizing the height.

Let $x$ represent the length of each side of the squares cut from the corners. When these squares are removed and the sides are folded up, the dimensions of the resulting box will be $12-2x$ by $12-2x$ by $x$.

The volume of the box can be expressed as $V(x) = x(12-2x)^2$.

To find the value of $x$ that maximizes $V(x)$, we take the derivative of $V(x)$ with respect to $x$, set it equal to zero, and solve for $x$.

$V'(x) = 12(12-2x)^2 - 2x(12-2x)(-2) = 0$

After simplifying and solving, we find $x = 3$.

Thus, the squares cut from the corners should be $3$ inches by $3$ inches each to maximize the volume of the box.