Question

Solve the following : An open box with a square base is to be made out of given quantity of sheet of area a². Show that the maximum volume of the box is `a^3/(6sqrt(3)`.

Answer 1

Let x be the side of square base and h be the height of the box.
Then x² + 4xh = a²

∴ h = `(a^2 - x^2)/(4x)`                    ...(1)

Let V be the volume of the box.

Then V = x²h

∴ V = `x^2((a^2- x^2)/(4x))`           ...[By (1)]

∴ V = `(1)/(4)(a^2x - x^3)`               ...(2)

∴ `"dV"/dx = (1)/(4)"d"/"dx"(a^2x - x^3)`

= `(1)/(4)(a^2 xx 1 - 3x^2)`

= `(1)/(4)(a^2 - 3x^2)`

and

`(d^2V)/(dx^2) = (1)/(4).d/dx(a^2 - 3x^2)`

= `(1)/(4)(0 - 3 xx 2x)`

= `-(3)/(2)x`

Now, `"dV"/dx = 0  "gives" (1)/(4)(a^2 - 3x^2)` = 0

∴ a² – 3x² = 0

∴ 3x² = a²

∴ x² = `a^2/(3)`

∴ x = `a/sqrt(3)`       ...[∵ x > 0]

and

`((d^2V)/dx^2)_("at" x = a/sqrt(3)`

= `-(3)/(2) xx a/sqrt(3)`

= `-sqrt(3)/(2) a < 0`

∴ V is maximum when x = `a/sqrt(3)`

From (2), maximum volume = `[1/4(a^2x - x^3)]_("at" x = a/sqrt3)`

= `(1)/(4)(a^2 xx a/sqrt(3) - a^3/(3sqrt(3)))`

= `(1)/(4)((2a^3)/(3sqrt(3)))`

= `a^3/(6sqrt(3)`

Hence, the maximum volume of the box is `a^3/(6sqrt(3)`  cu. unit.