An open box with a square base is to be made out of a given quantity of card board of area `c^2` square units. Show that the maximum volume of the box is `(c^3)/(6sqrt(3))` cubic units.

Explanation

Let the length of the square base of open box be x unit and is height be y units.
`therefore` Area of the metal used `=x^(2)+4xy`
`rArr y=(c^(2)-x^(2))/(4x)`……….(i) [given]
Now, volume of the box(V) =`x^(2)y`
`rArr V=x^(2). (c^(2)-x^(2))/(4x)`
`=1/4x(c^(2)-x^(2))`
`=1/4(c^(2)x-x^(3))`

On differentiating both sides w.r.t. x, we get
`(dV)/(dx) = 1/4(c^(2)-3x^(2))`.............(ii)
Now, `(dV)/(dx)=0 rArr c^(2)=3x^(2)`
`rArr x^(2)=c^(2)/3`
`rArr x^(2)=c^(2)/3`
`rArr x=c/sqrt(3)` [using positive sign]
Again, differentiating Eq. (ii) w.r.t. x, we get
`(d^(2)v)/(dx^(2))_("at " x=c/sqrt(3)) = -3/2.(c/sqrt(3))lt0`
Thus, we see that volume (V) is maximum at `x=c/sqrt(3)`.
`therefore` Maximum volume of the box, `(V)_(x=c/sqrt(3))=1/4(c^(2).c/sqrt(3)-c^(3)/(3sqrt(3)))`
`=1/4.(3c^(3)-c^(3))/(3sqrt(3))=1/4.(2c^(3))/(3sqrt(3))`
`=c^(3)/(6sqrt(3))` cu units