Let be the volume of the smallest right circular cone that can be circumscribed about a sphere of volume .
Find the ratio of to
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Let V c = 3 1 π R 2 H be the volume of the right circular cone and V s = 3 4 π r 3 be the volume of the sphere.
Let s = H 2 + R 2 be the slant height of the right circular cone.
From the geometry of the problem we have two similar right triangles. One with hypothenuse H − r and leg r and the other with hypotenuse s = H 2 + R 2 and leg R
⟹ r H − r = R H 2 + R 2 ⟹
R ( H − r ) = r H 2 + R 2 ⟹
R 2 ( ( H − r ) 2 − r 2 ) = H 2 r 2 ⟹
R 2 H ( H − 2 r ) = H 2 r 2 ⟹
R 2 = H − 2 r H r 2 ⟹
V c = π r 2 ( H − 2 r H 2 )
d H d V c = ( H − 2 r ) 2 π r 2 H ( H − 4 r ) = 0
( H < > 0 ⟹ H = 4 r and V c ( 4 r ) is a minimum since
2 r < H < 4 r ⟹ d H d V c < 0 and H > 4 r ⟹ d H d V c > 0
H = 4 r ⟹ R 2 = 2 r 2 ⟹ R = 2 r ⟹
V c = 3 8 π r 3 = 2 V s ⟹ V s V c = 2 .