Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is `tan^(-1)sqrt(2)`.

Explanation

Let 'r' be the radius of the base and 'h' be the height of cone.
` :. H^(2)+r^(2)= l^(2) ` where l is given slant height.
` rArr r^(2) = l^(2) - h^(2)` …(1)
Volume of cone ` V = 1/3 pi r^(2)h`
` rArr V= 1/3 pi h(l^(2)-h^(2))`
` 1/3 pi (l^(2)-h^(3))`
` rArr (dV)/(dh) = 1/3 pi (l^(2)-3h^(2))`
For maxima/minima, ` (dV)/(dh) = 0`
` rArr l^(2) - 3h^(2) =0 `
` rArr l^(2) = 3h^(2)` ...(2)
and ` (d^(2)V)/(dh^(2)) = 1/3 pi (0-6h) =- 2 pi h lt 0 `
` rArr ` V is maximum.
Now from equation (2)
` l^(2) = 3h^(2)`
` rArr r^(2) + h^(2) = 3h^(2)` [ from equation (1)]
` rArr r^(2) = 2h^(2)`
`rArr r=sqrt2*h rArr r/h = sqrt2`
` rArr tan theta = sqrt2`
Here ` theta` is the semi - vertical angle of cone
` rArr theta = tan^(-1) sqrt2`
Therefore, the semi-verticle angle of a cone of maximum volume is ` tan^(-1) sqrt2`.
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