QPaperGen

Let the side of the square cut off from each corner be \( x \,\text{cm} \).
\[\]
Then,
the length and breadth of the resulting box will each be \( (18 - 2x)\,\text{cm} \),
\[\] and the height will be \( x \,\text{cm} \).
\[\]
So, the volume \( V(x) \) of the box is:
\[
V(x) = x(18 - 2x)^2
\]

Differentiate:
\[
V'(x) = (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2)
= (18 - 2x)^2 - 4x(18 - 2x)
\]

Factor:
\[
V'(x) = (18 - 2x)[(18 - 2x) - 4x]
= (18 - 2x)(18 - 6x)
\]

Set derivative to zero:
\[
V'(x) = 0 \Rightarrow (18 - 2x)(18 - 6x) = 0
\Rightarrow x = 9 \text{ or } x = 3
\]

If \( x = 9 \), then \( 18 - 2x = 0 \), so length and breadth become zero.
This gives zero volume which is not acceptable.
\[\]

So, \( x = 3 \) is a critical point.
\[\]

Check second derivative:
\[
V''(x) = \frac{d}{dx}\left[(18 - 2x)(18 - 6x)\right]
= -2(18 - 6x) + (18 - 2x)(-6)
= -2(18 - 6x) - 6(18 - 2x)
\]

Evaluate at \( x = 3 \):
\[
V''(3) = -2(18 - 18) - 6(18 - 6) = -6 \cdot 12 = -72 < 0
\]

Since \( V''(3) < 0 \), the volume is \(\bf maximum\) at \( x = 3 \).

\[\]

Final Answer:
\[
\boxed{\text{The side of the square to be cut off is } 3\,\text{cm.}}
\]