Sphere in cone

A sphere of radius 3 cm desribe cone with minimum volume. Determine cone dimensions.

Result

r =  4.24 cm
h =  12 cm

Solution:

tanϕ=r/h tanϕ/2=3/r  V=13Sh=13πr2h V=13r3tanϕ V=13(r/tanϕ2)3tanϕ  V=9πtan3ϕ2(1cos2ϕ32tanϕtan1(ϕ/2)1cos2(ϕ/2)) V=0  cos2(ϕ/2)32tanϕtan1(ϕ/2)cos2ϕ=0 cos(ϕ/2)sin(ϕ/2)32cosϕsinϕ=0  12sinϕ32cosϕsinϕ=0 13cosϕ=0  ϕ=arccos13=1.2309594173=703144"  r=3/tan(703144"/2)=4.24 cm\tan \phi = r/h \ \\ \tan \phi/2 = 3/r \ \\ \ \\ V = \dfrac13 Sh = \dfrac13 \pi r^2h \ \\ V = \dfrac13 r^3 \tan \phi \ \\ V = \dfrac13 (r/\tan \dfrac{\phi}{2} )^3 \tan \phi \ \\ \ \\ V' = 9\pi \tan^{-3}\dfrac{\phi}{2}(\dfrac{1}{\cos^2 \phi} - \dfrac32 \tan \phi \tan^{-1}(\phi/2) \dfrac{1}{\cos^2 (\phi/2)} ) \ \\ V' = 0 \ \\ \ \\ \cos^2( \phi/2 ) - \dfrac32 \tan \phi \tan^{-1}(\phi/2) \cos^2 \phi = 0 \ \\ \cos ( \phi/2 ) \cdot \sin ( \phi/2 ) - \dfrac32 \cos \phi \sin \phi = 0 \ \\ \ \\ \dfrac12 \sin \phi - \dfrac32 \cos \phi \sin \phi = 0 \ \\ 1-3 \cos \phi = 0 \ \\ \ \\ \phi = \arccos \dfrac13 = 1.2309594173 = 70^\circ 31'44" \ \\ \ \\ r = 3/\tan(70^\circ 31'44"/2) = 4.24 \ \text{cm}
h=rtan703144"=12 cmh = r \tan 70^\circ 31'44" = 12 \ \text{cm}



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