Solving Lagrange System of Equations

Question

Given function $f(x,y)=x^2+xy+y^2$ on disc $x^2+y^2 = 1$

I get the LaGrange system of equations

$2x+y = 2xλ$

$2y+x = 2yλ$

$x^2+y^2 = 1$

I had many unsuccessful attempts but I remember being told that in general, setting all values of x,y,z to λ is the best way. How do you solve this problem to find all values x,y,z?

Answer 1

From the first two equation we get,

$$(2-2\lambda)x=-y$$

$$(2-2\lambda)y=-x$$

We may safely assume $x \neq 0$ (and) or $y \neq 0$ otherwise $(x,y)=(0,0)$ and the third equation $x^2+y^2=1$ is not satisfied.

Dividing both equations above by each other we must have $\frac{x}{y}=\frac{y}{x}$ or $x=\pm y$. If $x=\pm y$ then $x^2+x^2=1$ from the third equation so $x,y=\pm \frac{1}{\sqrt{2}}$.

Answer 2

Solve the first for $y$ and plug into the second. Note that $x$ cannot be $0$, so we can divide by it. $$2x(\lambda-1)=y\\2y(\lambda-1)=x\\4x(\lambda-1)^2=x\\ \lambda=\frac 32,\frac 12 \\x=\pm\frac {\sqrt 2}2, y=\pm\frac {\sqrt 2}2 \text { with the signs independent}$$