We are family - 2

Geometry Level 3

A family of ellipsoids is defined by the set of all ellipsoids whose equation is

x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1

and passing through the point ( 3 , 4 , 12 ) (3, 4, 12) . Find the smallest possible volume of an ellipsoid in this family.

288 3 π 288 \sqrt{3} \pi 576 3 π 576 \sqrt{3} \pi 4 3 ( 13 ) 3 π \dfrac{4}{3} (13)^3 \pi 144 2 π 144 \sqrt{2} \pi

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2 solutions

Mark Hennings
Jul 1, 2019

Applying the AM/GM inequality 1 3 = 1 3 ( 9 a 2 + 16 b 2 + 144 c 2 ) 9 × 16 × 144 a 2 b 2 c 2 3 = ( 144 a b c ) 2 3 \tfrac13 \; = \; \tfrac13\left(\tfrac{9}{a^2} + \tfrac{16}{b^2} + \tfrac{144}{c^2}\right) \; \ge \; \sqrt[3]{\tfrac{9 \times 16 \times 144}{a^2b^2c^2}} \; = \; \left(\tfrac{144}{abc}\right)^{\frac23} so the volume of the ellipsoid is V = 4 3 π a b c 4 3 π × 144 × 3 3 = 576 π 3 V \; = \; \tfrac43\pi abc \; \ge \; \tfrac43\pi \times 144 \times 3\sqrt{3} \; = \; \boxed{576\pi\sqrt{3}} We can choose a , b , c a,b,c to obtain equality in the AM/GM inequality, and hence can choose a , b , c a,b,c to obtain this minimum value of V V .

The volume of the ellipsoid becomes minimum when a=3√3, b=4√3 and c=12√3. The minimum volume is 4 3 \dfrac{4}{3} π(3√3)(4√3)(12√3)=576√3π

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