Side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 – 2a

Breadth, B = 18 – 2a and

Height, H = a

**Assuming, **

**Volume of box, **

V = LBH = a(18 - 2a)^{2}

**Condition for maxima and minima is**

\(\frac{dV}{da}\) = 0

⇒ (18 - 2a)^{2} + (a)(- 2)(2)(18 - 2a) = 0

⇒(18 - 2a)[(18 - 2a) - 4a] = 0

⇒ (18 - 2a)[18 – 6a] = 0

⇒ a = 3,9

\(\frac{d^2V}{da^2}\) = ( - 2)(18 – 6a) + ( - 6)(18 – 2a)

⇒ \(\frac{d^2V}{da^2}\) = 24a - 144

**For a = 3,**

\(\frac{d^2V}{da^2}\)** **=** **- 72,

⇒ \(\frac{d^2A}{d\theta^2}\) < 0

**For a = 9,**

\(\frac{d^2V}{da^2}\)** **= 72,

⇒ \(\frac{d^2A}{d\theta^2}\) > 0

**So for A to maximum,**

\(\frac{d^2A}{d\theta^2}\) < 0

**Hence, **

a = 3 will give maximum volume.

**And maximum volume, **

V = a(18 - 2a)^{2}

= 432 cm^{3}