Optimization Problem

Calculus Level 2

Let V c {\bf V_c } be the volume of the smallest right circular cone that can be circumscribed about a sphere of volume V s {\bf V_s} .

Find the ratio of V c {\bf V_c } to V s . {\bf V_s. }


The answer is 2.

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1 solution

Rocco Dalto
Oct 3, 2016

Let V c = 1 3 π R 2 H {\bf V_c = \frac{1}{3} \pi R^2 H } be the volume of the right circular cone and V s = 4 3 π r 3 {\bf V_s = \frac{4}{3} \pi r^3 } be the volume of the sphere.

Let s = H 2 + R 2 {\bf s = \sqrt{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 {\bf H - r } and leg r {\bf r } and the other with hypotenuse s = H 2 + R 2 {\bf s = \sqrt{H^2 + R^2} } and leg R {\bf R }

H r r = H 2 + R 2 R {\bf \implies \frac{H - r}{r} = \frac{\sqrt{H^2 + R^2}}{R} \implies }

R ( H r ) = r H 2 + R 2 {\bf R(H - r) = r\sqrt{H^2 + R^2} \implies }

R 2 ( ( H r ) 2 r 2 ) = H 2 r 2 {\bf R^2((H - r)^2 - r^2) = H^2 r^2 \implies }

R 2 H ( H 2 r ) = H 2 r 2 {\bf R^2 H (H - 2r) = H^2 r^2 \implies }

R 2 = H r 2 H 2 r {\bf R^2 = \frac{H r^2}{H - 2r} \implies }

V c = π r 2 ( H 2 H 2 r ) {\bf V_c = \pi r^2(\frac{H^2}{H - 2r}) }

d V c d H = π r 2 H ( H 4 r ) ( H 2 r ) 2 = 0 {\bf \frac{dV_c}{dH} = \frac{\pi r^2 H(H - 4r)}{(H - 2r)^2} = 0 }

( H < > 0 H = 4 r {\bf (H <> 0 \implies H = 4r } and V c ( 4 r ) {\bf V_c(4r) } is a minimum since

2 r < H < 4 r d V c d H < 0 {\bf 2r < H < 4r \implies \frac{dV_c}{dH} < 0 } and H > 4 r d V c d H > 0 {\bf H > 4r \implies \frac{dV_c}{dH} > 0 }

H = 4 r R 2 = 2 r 2 R = 2 r {\bf H = 4r \implies R^2 = 2r^2 \implies R = \sqrt{2}r \implies }

V c = 8 3 π r 3 = 2 V s V c V s = 2 . {\bf V_c = \frac{8}{3} \pi r^3 = 2V_s \implies \frac{V_c}{V_s} = \boxed{2}. }

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