Question

A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume

Answer 1

Let ‘x’ be the length of the box.

‘y’ be the height of the box.

Given, surface area = 108 sq.cm

i,e. 4(xy) + x² = 108

⇒ y = `(108 - x^2)/(4x)`

Volume of the box V = x²y

V = `x^2((108 - x^2)/(4x))`

V = `(108x - x^3/4`

`"dV"/("d"x) = (108 - 3x^2)/4`

Maximum or Minimum, 

`"dV"/("d"x)` = 0

⇒ 108 – 3x² = 0

x² = 36

x = ± 6 [x = – 6 is not possible]

∴ x = 6

Now, `("d"^2"V")/("d"x^2) = - (6x)/4 = - (3x)/2`

At x = 6

`("d"^2"V")/("d"x^2) < 0`

Volume of theboxis maximum when x = 6

y = `(108 - 36)/24`

= `72/24`

= 3

∴ Length of the box = 6 cm

Breadth of the box = 6 cm

Height of the box = 3 cm