Squares are cut from each corner of a #0.5m xx 0.5m# square of cardboard and the sides folded up to make an open box. What is the maximum possible volume of the box?

1 Answer
George C.
May 29, 2015

Consider the general case of a square sheet of cardboard with sides of length #s# and square cut outs at each corner with side #t#.

The area of the bottom of the resulting box will be #(s-2t)^2#.
The height of the box will be #t#.
The volume of the box will be #area xx height = (s-2t)^2*t#

#(s-2t)^2*t = (s^2-4st+4t^2)*t#

#= 4t^3-4st^2+s^2t#

The maximum of this will occur at some point where the derivative is zero...

#0 = d/(dt)(4t^3-4st^2+s^2t) = 12t^2-8st+s^2#

#=12t^2-6st-2st+s^2#

#=(12t^2-6st)-(2st-s^2)#

#=6t(2t-s)-s(2t-s)#

#=(6t-s)(2t-s)#

So #t=s/6# or #t=s/2#

If #t=s/2# the volume is #0# because #(s-2t) = 0#

So the maximum volume occurs when #t=s/6#.

Then

#volume = (s-2t)^2*t#

#= (s-2(s/6))^2*s/6#

#=((2s)/3)^2*s/6#

#=4/54s^3#

#=2/27s^3#

If #s=0.5m#

#s^3 = 0.125m^3#

and

#2/27s^3 = 1/108m^3 = 0.00dot(9)2dot(5)m^3 = 9259.dot(2)5dot(9)cm^3#

Related questions
Impact of this question
1755 views around the world
You can reuse this answer
Creative Commons License