Consider the general case of a square sheet of cardboard with sides of length ss and square cut outs at each corner with side tt.
The area of the bottom of the resulting box will be (s-2t)^2(s−2t)2.
The height of the box will be tt.
The volume of the box will be area xx height = (s-2t)^2*tarea×height=(s−2t)2⋅t
(s-2t)^2*t = (s^2-4st+4t^2)*t(s−2t)2⋅t=(s2−4st+4t2)⋅t
= 4t^3-4st^2+s^2t=4t3−4st2+s2t
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^20=ddt(4t3−4st2+s2t)=12t2−8st+s2
=12t^2-6st-2st+s^2=12t2−6st−2st+s2
=(12t^2-6st)-(2st-s^2)=(12t2−6st)−(2st−s2)
=6t(2t-s)-s(2t-s)=6t(2t−s)−s(2t−s)
=(6t-s)(2t-s)=(6t−s)(2t−s)
So t=s/6t=s6 or t=s/2t=s2
If t=s/2t=s2 the volume is 00 because (s-2t) = 0(s−2t)=0
So the maximum volume occurs when t=s/6t=s6.
Then
volume = (s-2t)^2*tvolume=(s−2t)2⋅t
= (s-2(s/6))^2*s/6=(s−2(s6))2⋅s6
=((2s)/3)^2*s/6=(2s3)2⋅s6
=4/54s^3=454s3
=2/27s^3=227s3
If s=0.5ms=0.5m
s^3 = 0.125m^3s3=0.125m3
and
2/27s^3 = 1/108m^3 = 0.00dot(9)2dot(5)m^3 = 9259.dot(2)5dot(9)cm^3227s3=1108m3=0.00.92.5m3=9259..25.9cm3
