25) Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \tan ^{-1} \sqrt 2

Answers (1)


Let a be the semi-vertical angle of cone
Let r , h , l are the radius , height , slent height  of cone
Now,
r = l\sin a \ and \ h=l\cos a
we know that
Volume of cone (V) = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l\sin a)^2(l\cos a) = \frac{\pi l^3\sin^2 a\cos a}{3}
Now,
\frac{dV}{da}= \frac{\pi l^3}{3}\left ( 2\sin a\cos a.\cos a+\sin^2a.(-\sin a)\right )= \frac{\pi l^3}{3}\left ( 2\sin a\cos^2a-\sin^3a \right )
\frac{dV}{da}=0\\ \frac{\pi l^3}{3}\left ( 2\sin a\cos^2a-\sin^3a \right ) = 0\\ 2\sin a\cos^2a-\sin^3a= 0\\ 2\sin a\cos^2a=\sin^3a\\ \tan^2 a = 2\\ a = \tan^{-1}\sqrt 2
Now,
\frac{d^2V}{da^2}= \frac{\pi l^3}{3}\left ( 2\cos a\cos^2a+2\cos a(-2\cos a\sin a+3\sin^2a\cos a) \right )
Now, at a= \tan ^{-1}\sqrt 2
\frac{d^2V}{dx^2}< 0
Therefore,  a= \tan ^{-1}\sqrt 2 is the point of maxima
Hence proved 
 

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