Minimise $2x+y$ subject to $x^2+y^2=1$ using KKT I'm doing this exercise in preparing for the final exam in optimization:

$$\begin{align*}
\text{min} &\quad 2x+y \\
\text{s.t} & \quad x^2+y^2=1
\end{align*}$$

Could you please verify if I correctly understanding the KKT's theorem? Thank you so much for your help! 

My attempt:
Let $f(x,y)=2x+y$ and $h(x,y)=x^2+y^2-1$. The problem becomes $$\begin{align*}
\text{min} &\quad f(x,y) \\
\text{s.t} & \quad h(x,y)=0
\end{align*}$$
Let $\mathcal K = \{(x,y) \in \mathbb R^2 \mid h(x,y)=0\}$. Then $\mathcal K$ is compact and $(0,0) \notin \mathcal K$. Moreover, $f$ is continuous. Hence the problem has at least one solution. All solutions $(\bar x,\bar y)$ are different from $(0,0)$.
We have $\mu \nabla h(\bar  x,\bar  y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \iff \mu \begin{pmatrix}2 \bar x \\2\bar  y \\ \end{pmatrix} = 0 \iff \mu =0$. So the constraint qualification is satisfied. Consider $$\begin{cases} \nabla f( x, y) + \mu \nabla h( x, y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\ { x}^2+{ y}^2-1=0 \end{cases} \iff \begin{cases} \begin{pmatrix}2 \\1 \\ \end{pmatrix} + \mu \begin{pmatrix}2 x \\2 y \\ \end{pmatrix} = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\{ x}^2+{ y}^2=1  \end{cases}$$
Solving this system, we get $$(\mu, x,  y) = (\sqrt{5}/2,-2/\sqrt{5} , -1/\sqrt{5})$$ and $$(\mu, x,  y) = (-\sqrt{5}/2,2/\sqrt{5} ,1/ \sqrt{5})$$
As such, $f(-2/\sqrt{5} , -1/\sqrt{5}) = -\sqrt{5}$ and $f(2/\sqrt{5} , 1/\sqrt{5}) = \sqrt{5}$.
Hence $(-2/\sqrt{5} , -1/\sqrt{5})$ is the solution to the problem.

Your final solution is correct. But this problem doesn't use inequalities in the constraint so KKT is not needed. Just basic Lagrangian multipliers. You should consider 
$$\begin{align*}
\text{max} &\quad xy \\
\text{s.t} & \quad x+y^2\leq2\\
&\quad x,y\geq 0
\end{align*}$$
Solution can be found here for your practice: http://www.math.ubc.ca/~israel/m340/kkt2.pdf
