Show that the right circular cone of least curved surface and given volume has an altitude equal to `sqrt(2)`time the radius of the base.

Explanation

Let 'r' be radius and 'h' be the height of the conical tent.
If the slant height is l then,
`l^(2)= h^(2)| r^(2)` ….(1)
Volume ` V = 1/3 pi r^(2) h ` …(2)
Now, curved surface ` C = pi r l= pi r sqrt(h^(2)+r^(2))`
= ` pi r sqrt(((3V)/(pi r^(2)))+r^(2)`
[From equation (1)]
` rArr C^(2) = pi^(2)r^(2)((9V^(2))/(pi^(2)r^(4))+r^(2))`
Let ` Z= C^(2)=(9V^(2))/r^(2) + pi^(2) r^(4)`
` rArr (dZ)/(dr) = (-18V^(2))/r^(3) + 4pi^(2)r^(3)`
and ` (d^(2)Z)/(dr^(2)) = (54V^(2))/r^(4) + 12pi^(2)r^(2)`
For maxima/minima, ` (dZ)/(dr) = 0 `
` rArr (-18V^(2))/r^(3) + 4pi^(3)r^(3)= 0`
` rArr 18V^(2) = 4 pi^(2)r^(6)`
` 18*1/9 pi^(2)r^(4)h^(2) = 4pi^(2)r^(6)`
` rArr 2h^(2) 2h^(2) = 4r^(2) rArr h = sqrt2r`
` rArr h/r = sqrt2/1 = sqrt(2 ): 1`
and ` (d^(2)Z)/(dr^(2)) gt 0 `
` rArr ` Z is minimum.
` rArr C^(2)` is minimum.
` rArr ` C is minimum.
Therefore, minimum canvas is required in tent if the ratio of its height and radius of base is ` sqrt(2) : 1` .