Let T be a tangent plane to the ellipsoid x^{2}/a^{2} + y^{2}/b^{2} + z^{2}/c^{2} = 1. What is the smallest possible volume for the tetrahedral volume bounded by T and the planes x = 0, y = 0, z = 0?

**Solution**

Answer: √3 abc/2.

The normal at (x_{0}, y_{0}, z_{0}) is (xx_{0}/a^{2}, yy_{0}/b^{2}, zz_{0}/c^{2}). So the tangent plane is x x_{0}/a^{2} + y y_{0}/b^{2} + z z_{0}/c^{2} = 1. This cuts the three axes at x = a^{2}/x_{0}, y = b^{2}/y_{0}, z = c^{2}/z_{0}. We can regard two of these lengths as defining the base of the tetrahedron, and the third as forming its height. Hence its volume is a^{2}b^{2}c^{2}/(6x_{0}y_{0}z_{0}).

We wish to maximize x_{0}y_{0}z_{0}. That is equivalent to maximising x_{0}^{2}/a^{2} y_{0}^{2}/b^{2} z_{0}^{2}/c^{2}. But we know that the sum of these three numbers is 1, so their maximum product is 1/27 (achieved when they are all equal - the arithmetic/geometric mean result). Hence x_{0}y_{0}z_{0} has maximum value abc/(3√3).

© John Scholes

jscholes@kalva.demon.co.uk

14 Dec 1999