Let the side of square be x and ` (0 lt x lt 9)` , then the length of box is (18 - 2x), breadth is (18-2x) and height is x . Let V be the volume of the open box formed by folding the flaps of tin.

`:. V = x ( 18 - 2x)(18-2x)`

` = 4x(9-x)^(2)=4x(81+x^(2)-18x)`

` = 4(x^(3)-18x^(2)+81x)`

` rArr(dV)/(dx) = 4(3x^(2)-35x+81)= 12 (x^(2)-12x+27)`

and ` (d^(2)V)/(dx^(2)) = 12 (2x- 12) = 24 (x - 6) `

For maximum value

` (dV)/(dx) = 0 rArr 12 (x^(2)- 12x + 27) = 0`

` rArr x ^(2) - 12x + 27 = 0`

` rArr ( x - 3 ) (x - 9) = 0`

` rArr x = 3, 9 `

but x = 9 is impossible so x = 3.

at ` x= 3 , ((s^(2)V)/(dx^(2)))_(x = 3 ) = 24 (3 - 6)=- 72 lt 0 `

` :. x = 3 `

` rArr ` Side of square cut = 3 cm