Question

Show that the right circular cone of least curved surface and given volume has an altitude equal to `sqrt2` time the radius of the base.

Answer 1

Let the radius of the cone = r

Height of the cone = h

`therefore V = 1/3 pir^2 "h"` = constant quantity

`therefore r^2h = (3 xx "constant quantity")/pi = k` (assumed)

`r^2h = k, h = k/r^2`                ...(1)

Curved surface S = `pirl = pir sqrt(h^2 + r^2)`

S = `pir sqrt(k^2/r^4 + r^2)`

`= pir sqrt((k^2 + r^6)/r^4)`

`= pi/r sqrt(k^2 + r^6)`

On differentiating,

`therefore (dS)/(dr) = pi [((6r^5)/(2sqrt(r^6 + k^2)) xx r - sqrt(r^6 + k^2) * 1)/r^2]`

`= pi * (3r^6 - (r^6 + k^2))/(r^2 sqrt(r^6 + k^2))`

`= (2r^6 - k^2)/(r^2 sqrt(r^6 + k^2))`

For maximum and minimum, `(dS)/(dr) = 0`

`=> 2r^6 - k^2 = 0`

`=> r^6 = k^2/2`

`=> r^6 = (h^2 r^4)/2`          ...(2)

⇒ h² = 2r²

∴ h = `sqrt2 r`

At h = `sqrt2 r`, as r passes through `sqrt2 r`

`(dS)/(dr)` changes from -ve to + ve.

∴ S is minimum when h = `sqrt2 r`

Therefore, the height of the right circular cone with the minimum curved surface is `sqrt2` times the radius.