A square piece of tin of side 18 cm is to be made into a box without top, by cutting-off square from each corner and folding up the flaps of the box. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Let the side of the square to be cut-off be x cm (0<x<9). Then, the length and the breadth of the box will be (18-2x) cm each and the height of the vbox is x cm

Let V the volume of the open box formed by folding up the flaps, then
$V=x(18-2x)(18-2x)=4x(9-x{)}^2=4x(81+x^2-18x)=4(x^3-18x^2+81x)$
On differentiating twice w.r.t. x, we get
$\frac{dV}{dx}=4(3x^2-36x+81)=12(x^2-12x+27)$
and $\frac{d^{2}V}{d{x}^2}=12(2x-12)=24(x-6)$
For maxima put $\frac{dV}{dx}=0\Rightarrow 12(x^2-12x+27)=0$
$\Rightarrow x^2-12x+27=0\Rightarrow (x-3)(x-9)=0\Rightarrow x=3,9$
But x=9 is not possible.
$\because 2x=2\times 9=18$
At $x=3{\frac{d^{2}V}{d{x}^2}}_{x=3}=24(3-6)=-72<0$
$\therefore$ By second derivative test, x=3 is the point of maxima.
Hence, if we cut off the side 3 cm 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.