Lagrange multiplier problem - how to solve x, y and lambda?

Question

A typical Lagrange multiplier problem. Have I done right and how to continue?

$ L(x,y,\lambda) = x(1-y)+\lambda (x^2+y^2-1) $

$\Rightarrow$

$ L_x = 1-y + 2\lambda x=0$

$ L_y = -x + 2\lambda y=0$

$ L_\lambda = x^2+y^2 - 1 =0$

How to solve $x$, $y$ and $\lambda$?

Answer 1

Note that if $y=0$ then $x=0$ which is illegal. So then get rid of $\lambda$ in the two equations $$ 1+2\lambda x=y\\ x=2\lambda y\\ \implies \frac{x}{y}=2\lambda\implies y=1+\frac{x}{y}x\implies y^2=y+x^2 $$ now get $x^2=1-y^2$ involved to find $$ y^2=y+1-y^2\implies 2y^2-y-1=0\implies (2y+1)(y-1)=0 $$ and $y=1$ or $y=-\frac12$. If $y=-\frac12$ we have $$ x=\pm\frac{\sqrt{3}}{2} $$ if $y=1$, we have $x=0$.

Answer 2

We consider

$$( F- \lambda G)$$

$$ \dfrac{F}{G}= \dfrac{F_x}{G_x}= \dfrac{F_y}{G_y}=\lambda $$

$$ \dfrac{1-y}{2x}= \dfrac{-x}{2y}= \lambda $$

simplifying we have to solve

$$ x^2-y^2+y=0,\quad x^2+y^2-1=0,\,$$

subtracting

$$ 2 y^2-y-1=0 ,\rightarrow \,y= (1,-\frac12) $$

so $(x,y)$ results in:

$$ (x,y)= (0,1), (\pm \sqrt3/2,-\frac12)$$

and Lagrange multipliers $\lambda$ from second equation calculate to $ \pm \sqrt{3}/2 $

It is to be noted there are three critical points.

Area is maximized as shown yellow, unit circle constraint boundary is geometrically depicted below hopefully for a comprehensive understanding,