Answer :

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.

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