#### 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.

#### Solution

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}

\[V'\left( x \right) = \left( 18 - 2x \right)^2 - 4x\left( 18 - 2x \right)\]

\[ = \left( 18 - 2x \right)\left( 18 - 2x - 4x \right)\]

\[ = \left( 18 - 2x \right)\left( 18 - 6x \right)\]

\[ = 12\left( 9 - x \right)\left( 3 - x \right)\]

\[V''\left( x \right) = 12\left( - \left( 9 - x \right) - \left( 3 - x \right) \right)\]

\[ = - 12\left( 9 - x + 3 - x \right)\]

\[ = - 24\left( 6 - x \right)\]

\[\text { For maximum and minimum values of V, we must have }\]

\[ V'\left( x \right) = 0\]

\[\Rightarrow\] *x* = 9 or *x* = 3

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

*∴ x* ≠ 9

*\[\Rightarrow\] x *= 3

Now,

\[V''\left( 3 \right) = - 24\left( 6 - 3 \right) = - 72 < 0\]

*∴ x* = 3 is the point of maxima.

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

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 cm^{3}