Question

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.

Answer 1

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.