What is the minimum volume bounded by the planes $x=0, y=0, z=0$ and a plane which is tangent to the ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1$$ where $x,y,z>0$
I only know that when we want to find maximum and minimum, there is constraint $g$ such that $\nabla f = a\nabla g$ where $a$ is Lagrange multiplier.
What is the constraint and what is $f$?
Best Answer
Here is an illustration of the problem at hand.
You want to find the the volume of the tetrahedron formed by the $x=y=z=0$ planes, and the plane tangent to the ellipsoid at some point on its surface, $(x,y,z)$. This volume, of course, will vary with different points on the surface of the ellipse, so the constraint function must be the following equality
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Now the vector normal to the surface of the ellipsoid at some point $(x,y,z)$ is given by the gradient of its equation.
$$\nabla g (x,y,z) = \left<\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}\right>$$
The equation for the plane is tangent to the ellipsoid at this point is easily worked out to be
$$\frac{x}{a^2}x'+\frac{y}{b^2}b'+\frac{z}{c^2}z'=1$$
Now, finally, for the volume that this plane bounds within the first quadrant, we make use of the fact that this volume is simply $1/6$ the volume of the rectangular prism whose side lengths are given by the points of where the plane intersects the three axes.
$$V=\frac{(abc)^2}{6xyz}$$
And that is your function $f$ which you are trying to minimize.
$$f=V$$
$$g\rightarrow \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$