QPaperGen

Let the side of each square cut off be \( x\,\text{cm} \).

- After folding, the height of the box = \( x \)
- Length of the box = \( 45 - 2x \)
- Breadth of the box = \( 24 - 2x \)

So, the volume \( V(x) \) of the box is:
\[
V(x) = x(45 - 2x)(24 - 2x)
\]

Expand:
\[
V(x) = x[(45)(24) - 2x(24) - 2x(45) + 4x^2]
= x[1080 - 48x - 90x + 4x^2]
= x[1080 - 138x + 4x^2]
= 4x^3 - 138x^2 + 1080x
\]

Differentiate:
\[
V'(x) = 12x^2 - 276x + 1080
\]

Factor:
\[
V'(x) = 12(x^2 - 23x + 90)
= 12(x - 18)(x - 5)
\]

Set derivative to zero:
\[
V'(x) = 0 \Rightarrow x = 18 \text{ or } x = 5
\]

Now, \( x = 18 \) is not valid since:
- \( 2x = 36 > 24 \) → length or breadth becomes negative or zero.

So, only possible value is \( x = 5 \).

Check second derivative:
\[
V''(x) = \frac{d}{dx}(12x^2 - 276x + 1080) = 24x - 276
\]

Evaluate at \( x = 5 \):
\[
V''(5) = 24(5) - 276 = 120 - 276 = -156 < 0
\]

So, \( x = 5 \) is a point of **local maximum**.

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### **Final Answer**:
\[
\boxed{\text{The side of the square to be cut off is } 5\,\text{cm.}}
\]