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 cm3