Lagrange Multipliers Calculus II Question It is given me that $f(x,y)=x^2+-x+2y^2$ subject to $g(x,y)=x^2+y^2=1$ and asks for maximum and/or minimum.
What I did...
Equalize the partial derivatives and add the function $g$ to the system:
$$2x-1=2x\gamma$$
$$4y=2y\gamma$$
$$x^2+y^2=1$$
Then on the second equation I canceled $4y/2y$ to find $\gamma=2.$
After that solved the remaining equations to find two points $(-\frac{1}{2},\frac{\sqrt 3}{2})$ and $(-\frac{1}{2},-\frac{\sqrt 3}{2})$.
The answer that was given is that $(1,0)$, $(-1,0)$ are local minimum and $(-\frac{1}{2},\frac{\sqrt 3}{2})$, $(-\frac{1}{2},-\frac{\sqrt 3}{2})$ are local maximum.
My question is where the hell did that $(1,0)$ and $(-1,0)$ came from?

From $4y = 2y\gamma$, if $y = 0$ then $\gamma$ isn't specified. In which case from $x^2 + y^2 = 1$, we have $x = \pm 1$. I.e., the points $(1,0)$ and $(-1,0)$, which correspond respectively to $\gamma = 1/2$ and $\gamma = -3/2$.
