Lagrange System of Equations

Question

I am having hard time solving this system of equations

$2x+3λx^2-3λyz=0$

$2y+3λy^2-3λxz=0$

$2z+3λz^2-3λxy=0$

$x^3+y^3+z^3-3xyz=1$

I managed to find λ:

$λ=\frac{2x}{3yz-3x^2}$

$λ=\frac{2y}{3xz-3y^2}$

$λ=\frac{2z}{3yx-3z^2}$

Any tips?

Answer 1

your system is equivalent to finding the closest point(s) to the origin on $x^3 + y^3 + z^3 - 3xyz = 1.$

The surface is actually a surface of revolution around the ray $x=y=z,$ $x+y+z > 0.$ As a result, there is an entire circle of points at the minimum distance to the origin, occurring in some specific plane $x+y+z = D$ Parametrizing the curve is the reason Wolfram gives $y$ and $z$ as functions of $x$

You can calculate this using purely rotated coordinates $$ u = \frac{x+y+z}{\sqrt 3} \; , $$ $$ v = \frac{-x+y}{\sqrt 2} \; , $$ $$ w = \frac{-x-y+2z}{\sqrt 6} \; . $$ The surface is $$ u = \frac{ \left( \frac{2}{\sqrt {27}} \right)}{ \; \; \; v^2 + w^2 \; \; \; } $$

Think of the vertical axis below as the 1,1,1 ray, while rotating the surface around that ray