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Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin inverse 1 by 3

26) Show that semi-vertical angle of the right circular cone of given surface area and
maximum volume is \sin ^{-1} (1/3)

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Let r, l, and h are the radius, slant height and height  of cone respectively
Now,
r = l\sin a \ and \ h =l\cos a
Now,
we know that
The surface area of the cone (A) = \pi r (r+l)
A= \pi l\sin a l(\sin a+1)\\ \\ l^2 = \frac{A}{\pi \sin a(\sin a+1)}\\ \\ l = \sqrt{\frac{A}{\pi \sin a(\sin a+1)}}
Now, 
Volume of cone(V) =

 \frac{1}{3}\pi r^2h = \frac{1}{3}\pi l^3 \sin^2 a\cos a= \frac{\pi}{3}.\left ( \frac{A}{\pi\sin a(\sin a+1)} \right )^\frac{3}{2}.\sin^2 a\cos a
On differentiate it w.r.t to a and after that
\frac{dV}{da}= 0
we will get 
a = \sin^{-1}\frac{1}{3}
Now, at  a = \sin^{-1}\frac{1}{3}
\frac{d^2V}{da^2}<0
Hence, we can say that a = \sin^{-1}\frac{1}{3} is the point if maxima 
Hence proved

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