Question

An open box is to be made out of a piece of a square card board of sides 18 cms by cutting off equal squares from the comers and turning up the sides. Find the maximum volume of the box.

Answer 1

Let the side of the square cut off from the corners be x cm.

Therefore, each side of the square box is (18 – 2x) cms and the height is x cms.

Let V be the volume of the box.

V = Area of the base × Height

V = (18 − 2x)²x 

V = (324 − 72x + 4x²) x

∴ V = 4x³ − 72x² + 324x

Differentiating w.r.t. x, we get

`(dV)/dx = 12x^2 - 144x + 324`

`therefore (d^2V)/dx^2 = 24x - 144`

For maximum volume, `(dV)/dx = 0`

∴ 12x² − 144x + 324 = 0

∴ x² − 12x + 27 = 0

∴ (x − 3) (x − 9) = 0

∴ x − 3 = 0 or x − 9 = 0

∴ x = 3 or x = 9

But x ≠ 9

∴ x = 3

For x = 3

`((d^2V)/dx^2)_(x = 3) = 24(3) - 144 = -72 < 0`

The volume of the box is maximum when x = 3.

∴ Maximum value of the box = (18 − 6)² (3) 

= 432 cc