Lagrange Multipliers Calculus II Question

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?

Answer 1

If $y=0$ then we cannot cancel it. Moreover, $x^2+y^2=1$ implies $x= \pm 1$ when $y=0$ hence the mysterious extra points.

Answer 2

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 $\gamma = -3/2$.