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Question
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is `tan^(-1) sqrt(2)`
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Solution
If θ is the semi-vertical angle and l is the given slant height, then radius of base
= l sin θ, height = l cos θ ... (∵ ABC is right-angled triangle)
and volume of cone = `1/3 pir^2h`
⇒` V = 1/3 pi (l sin theta)^2 lcos theta 1/3 pil^3 sin^2 theta costheta`
Where V be the volume.
`(dV)/(d theta) = 1/3 pil^3 {(sin^2 theta) (- sin theta) + cos theta xx 2 sin theta cos theta}`
`= 1/3 pil^3 sin theta [-sin^2 theta + 2 (1 - sin^2 theta)]`
`= 1/3 pil^3 sin theta cos^2 theta [2 sec^2 theta - 3 tan^2 theta]`
`= 1/3 pil^3 sin theta cos^2 theta [2 - tan^2 theta]`
For maximum / minimum volume, let `(dV)/(d theta) = 0`
`= 1/3 pil^3 sin theta cos^2 theta (2 - tan^2 theta) = 0`
`= tan theta = sqrt 2`
`= theta = tan^-1 sqrt2`
`= (d^2V)/(d theta)^2 = 1/3 pil^3 cos^3 theta (2 - 7 tan^2 theta)`
`= ((d^2V)/(d theta^2))_(tan theta= sqrt2)`
`= 1/3 pi l^3 (1/sqrt3)^3 (2 - 7 xx 2)`
`= (4pil^3)/(3sqrt3) < 0`
Thus, V is maximum when
`tan theta = sqrt 2 or theta = tan^-1 sqrt 2`
i.e., when the semi - vertical angle of the cone is `tan ^-1 sqrt2`.
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