Let,
x= length of cut on each sides of small squares
V= volume of folded box
The length of base after two cuts from each edge of size
x is
(12−2x) and depth of box after folding is
x. Two equal length cut from
12 m can each have a maximum length of
6 m. So the value of
x lies between 0 and 6 and at that values of
x volume is zero.
Volume of box can be written as
V=x(12−2x)(12−2x)=4x3−48x2+144x To find out the maximum value we have to find the critical point,
dVdx=0 dVdx=12x2−96x+144=12(x−2)(x−6)=0 x=2,6 There is only one value
x=2 is possible at which
V=128 m3 and this is the maximum volume of box
because the volume of box is zero at
x=6.