Minimun volume of a tetrahedron bounded by an ellipsoid and tangent plane

Question

Let P be a tangent plane to the ellipsoid $$ (x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1 $$ at a point in the first octant. Let T be a tetrahedron bounded by P and the coordinate planes x=0, y=0, z=0. Find the minimum volume of T. (The volume of a tetrahedron is​ one-third the area of the base times the​ height.)

I would show the work I have for this problem, but I really do not know where to start.

Answer 1

let $u = \frac {x}{a}, v = \frac {y}{b}, w = \frac {z}{c}$

$u^2 + v^2 + w^2 = 1$

The tetrahedron that minimizes volume will be bound by $u+v+w = \sqrt 3$

$V = \frac 16 (\sqrt 3)^3 = \frac {\sqrt 3}{2}$

But this volume is in $uvw$ space. And $uvw$ space is compressed by a factor of $abc$

$V = \frac {\sqrt 3}{2} abc$

Alternatively, using calculus.

$\nabla (\frac {x^2}{a^2}+\frac {y^2}{b^2}+\frac {z^2}{c^2} = 1) = (\frac {2x}{a^2},\frac {2y}{b^2},\frac {2z}{c^2})$

For some point $(x_0,y_0,z_0)$ on the surface of the ellipsoid the equation of plane $P$ is

$\frac {x_0}{a^2}x+\frac {y_0}{b^2}y + \frac {z_0}{c^2}z = 1$

Which intercepts the coordinate axes at:

$(\frac {a^2}{x_0}, 0,0),(0,\frac {b^2}{y_0}, 0).(0,0, \frac {c^2}{z_0})$

and $V = \frac 16 \frac {(abc)^2}{x_0y_0z_0}$

Objective: Minimize $xyz$ constrained by $\frac {x^2}{a^2}+\frac {y^2}{b^2}+\frac {z^2}{c^2} = 1$

Lagrange multipliers

$yz = 2\lambda \frac{x}{a^2}\\ xz = 2\lambda \frac{y}{b^2}\\ xy = 2\lambda \frac{z}{c^2}$

$\frac {a^2yz}{x} = \frac {b^2xz}{y} =\frac {c^2xy}{z} = 2\lambda\\ \frac {x^2}{a^2} = \frac {y^2}{b^2} = \frac {z^2}{c^2}$

And by the constraint of

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

gives us

$\frac {x^2}{a^2} = \frac {y^2}{b^2} = \frac {z^2}{c^2} = \frac 13$

$\frac {xyz}{abc} = (\sqrt {\frac 13})^3\\ \frac 16 \frac{(abc)^2}{xyz} = \frac {\sqrt {3}}{2}(abc)$

Wasn't it easier without calculus?

Answer 2

Since this is about volume, and not about lengths or areas, we may assume $a=b=c=1$. Let $P=(u,v,w)\in S^2$. The tangent plane to $S^2$ at $P$ is then given by $$ux+vy+wz=1$$ and cuts the axes at ${1\over u}$, ${1\over v}$, ${1\over w}$. The volume of the tetrahedron is minimal when $uvw$, or $u^2v^2w^2$, is maximal under the constraint $u^2+v^2+w^2=1$. By the AGM inequality this is the case iff $u=v=w={1\over\sqrt{3}}$. We therefore obtain $${\rm vol}(T_{\rm opt})={1\over6}(\sqrt{3})^3={\sqrt{3}\over2}\ .$$ Multiply by $abc$ for general $a$, $b$, $c$.