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

Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be (18 − 2x) cm each and height of the box will be x cm.

Volume of the box, V(x) = x(18 − 2x)²

$$\begin{gathered} V\text{'}\left(x\right)={\left(18-2x\right)}^2-4x\left(18-2x\right) \\ =\left(18-2x\right)\left(18-2x-4x\right) \\ =\left(18-2x\right)\left(18-6x\right) \\ =12\left(9-x\right)\left(3-x\right) \\ V\text{'}\text{'}\left(x\right)=12\left(-\left(9-x\right)-\left(3-x\right)\right) \\ =-12\left(9-x+3-x\right) \\ =-24\left(6-x\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For}\mathrm{maximum}\mathrm{and}\mathrm{minimum}\mathrm{values}\mathrm{of}V,\mathrm{we}\mathrm{must}\mathrm{have} \\ V\text{'}\left(x\right)=0 \end{gathered}$$
$\Rightarrow$x = 9 or x = 3

If x = 9, then length and breadth will become 0.

x ≠ 9

x = 3

Now,
$V\text{'}\text{'}\left(3\right)=-24\left(6-3\right)=-72<0$

x = 3 is the point of maxima.

$V\left(x\right)=3{\left(18-6\right)}^2=3\times 144=432{\mathrm{cm}}^3$

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box so obtained would be the largest, i.e. 432 cm³.