A square piece of

Let the side of the square to be cut off be x, then, the length and the breadth of the box will be (18 – x) cm each and the height of the box is x cm.

Then, the volume {V(x)} of the box is given by:

V(x) = x(18-x)²

⇒ V’(x) = (18-x)² - 2x(18-x)

= (18 - x)[18- x -2x]

= (18 - x)(18 - 3x)

Now, V’’(x) = (18 - x)(-3) + (18 - 3x)(-1)

= -3(18 - x) - (18 - 3x)

= -54 + 3x - 18 + 3x

= 6x - 72

Now, V’(x) = 0

⇒ x = 18 or 3

If x = 18 then breadth becomes 0 which is not possible

Therefore, x =3

V’’(3) = 6.3 - 72 = -ve

Then, by second derivative test, x= 3 is the point of maxima of V.

Therefore, If we remove a square of side 3cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.