Constrained Extrema with Lagrange Let $f(x,y) = x^2 +y^2$  with $ g(x,y) = 6x^2+4xy+9y^2-50$.
$\Longrightarrow(2x,2y) + \lambda \cdot (12x+4y,18y+4x)$
So I have to solve $$2x + 12x \lambda +4y \lambda = 0,$$
$$2y+4x\lambda+18y\lambda=0.$$
I am not really sure on how to proceed here. Do I try different values for $x$ or $y$ which are acceptable under the restriction $g(x,y)$ ?
I tried to multiply the first eq. with $y$ and the second with $x$ but didn't get far in solving that...
 A: Outline:

(1) Regarding $x,y$ as unknowns, and $\lambda$ as an unknown constant, you have a homogeneous system of two linear equations in two unknowns.

(2) The solution $(x,y)=(0,0)$ fails to satisfy $g(x,y)=0$, so can be rejected. For the system to have a solution other than $(x,y)=(0,0)$, the two equations must be linearly dependent. This yields an equation for $\lambda$ (a $2 \times 2$ determinant must be $0$). Solve that equation.

(3) For each value of $\lambda$, substitute that value into either of the two equations, and solve for $y$ in terms of $x$, or $x$ in terms of $y$, whichever is more convenient.

(4) Substitute the result of step (3) into the constraint $g(x,y)=0$, and solve for $x^2$ or $y^2$ (whichever is present).

(5) Using the results of step (3), solve for the other of $x^2,y^2$.

(6) For each candidate pair $(x^2,y^2)$, evaluate $f(x,y)$, and compare.
A: We have
\begin{eqnarray*}
x+6x \lambda +2 y \lambda=0 \\
y+9y \lambda + 2 x \lambda=0
\end{eqnarray*}
Rearrange the first equation to $x= \frac{-2y \lambda}{1+6 \lambda}$ and substitute this into the second equation. We get
\begin{eqnarray*}
1+6 \lambda +9 \lambda (1+6 \lambda)-4 \lambda^2=0 \\
50 \lambda^2 + 15 \lambda+1=0.
\end{eqnarray*}
Reckon you can do it from here ?
