Find the maximum and minimum values of $x^2+y^2+z^2$ subject to the condition $ax^2+by^2+cz^2 =1\;\;$ and $\;\;lz+my+ny=0$ and interpret the result geometrically
I started with Lagrange's method and Let $F = x^2+y^2+z^2+\lambda(lz+my+ny)\;\;$ . I found $\frac{\partial F}{\partial x} = F_x = 2x+\lambda(l)=0\;\;$ therefore $x=-(\lambda*l)/2\;\;$ and similarly I found $y,z\;\;$ . I substituted these values in $ax^2+by^2+cz^2 =1\;\;for \;\lambda$ and therefore $\lambda = \frac{2}{\sqrt{al^2+bm^2+cn^2}}$ .
Therefore $x=\frac{-l}{\sqrt{\sum al^2}},$ $y=\frac{-m}{\sqrt{\sum al^2}}$ $z=\frac{-n}{\sqrt{\sum al^2}}$ and the Max value of $\;\;x^2+y^2+z^2\;\;$ is $\frac{l^2+m^2+n^2}{\sqrt{al^2+bm^2+cn^2}}$ .
But when I substitute the values of $x,y,z$ back into $\;\;lz+my+ny$ I am not getting zero, therefore I made a mistake somewhere . Can someone give the solution and also the geometric interpretation.
You have two constraints, so you should have two Lagrange multipliers. Look at the example at the bottom of this page from Paul's online math notes.
The geometric interpretation is easy. The constraints describe a quadric surface and a plane, so we are looking for a point on their intersection. Since we are optimizing $x^2+y^2+z^2,$ we are looking for the maximum and minimum distances from that intersection curve to the origin.