Solving a system of equations coming from Lagrange multiplier method

Question

I'm trying to use Lagrange multipliers to find the points on a surface given by $g(x,y) = x^4+y^4-4xy = 6$ in $R^2$ which have the smallest and largest distance to the origin $(0,0)$. So $g(x,y)=6$ is my constraint, and the function I am minimizing is the distance $d(x,y) = \sqrt{x^2+y^2}$ which is equivalent to minimizing $f(x,y)=x^2+y^2$. So by the Lagrange multiplier method the system I have to solve is $$ 2x = \lambda(4x^3-4y) $$ $$ 2y = \lambda(4y^3-4x) $$ $$ x^4+y^4-4xy = 6 .$$ The problem I'm having is solving this system for $\lambda$. I have tried to use the suggestion in this question (Solve system of linear equations within a Lagrange multiplier problem) and setting $y=xt$ (I'm not sure why we're even allowed to do this in the first place?). Plugging this into the first two equations and then dividing one by the other leads me to the quadratic $(x^2+1)t^2-x^2t-1 = 0$ in the variable $t$, which has solutions $$ t = \frac{x^2 \pm \sqrt{x^4+4(x^2+1)}}{2(x^2+1)} $$ but this looks extremely messy and I'm not sure if it helps me to solve the original system? Am I on the right track, or is there any easier way?

Answer 1

HINT.-You have $\dfrac{x}{2x^3-2y}=\dfrac{y}{2y^3-2x}\Rightarrow (x^2-y^2)(1-xy)=0$. You can take now $x=y$ which gives $x^4-2x^2-3=0$ so $x=\sqrt3$ and $$\lambda=\dfrac{x}{2x^3-2y}=\frac{\sqrt3}{18\sqrt3-2\sqrt3}=\frac{1}{16}$$ Complete the analysis with the other possible values.