Question

Find the point on an given ellipsoid that is the farthest to a given surface.(distance between point on ellipsoid and plane should be max)... Should I use Lagrange multipliers?

Answer 1

\[ D(x) = \frac{|px+qy+zr + d|}{\sqrt{p^2+q^2+r^2}}\]
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]

Not really sure how to do it now!!

Answer 2

The equations are:
ellipsoid: \[\left( x -3 \right)^{2}/3 + y^2/4+z^2/5=1\]
surface: \[3x+4y^2+6z+6=0\]
It would be so much easier that the second equation is equation of a plane.

Answer 3

still there's a problem ... how do you find the projection of point to a surface

Answer 4

looks like you need to find the distance first ... using normal and dummy point (x1,y1.z1)

r = (x1,y1,z1) + t(3,4y1,6)
find the distance in in terms of x1,y1,z1 ... which would a function under constraint of ellipse ...
then use lagrange multipliers ...
not sure if this will work out!!!