<s>[INST] 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.
 [/INST] 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
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