Minimize $x+2y$ subject to $x^2+y^2\le1$ and $3x+4y\le-5$. I want to minimize $x+2y$ subject to $x^2+y^2\le1$ and $3x+4y\le-5$.
I found the gradient conditions: $1+2x\lambda_1+3\lambda_2=2+2y\lambda_1+4\lambda_2=0$.
Then, by complementary slackness: $\lambda_1(x^2+y^2-1)=\lambda_2(3x+4y+5)=0$.
Wolfram-Alpha gives $(x,y,\lambda_1,\lambda_2) = (\frac1{\sqrt5},\frac2{\sqrt5},-\frac{\sqrt5}2,0)$ or $(-\frac1{\sqrt5},-\frac2{\sqrt5},\frac{\sqrt5}2,0)$. The second one gives an objective value of $-\sqrt5$, but both of these violate the second constraint.
The actual mimimum is $-\frac{11}5$ at $(x,y) = (-\frac35,-\frac45)$. Where did I go wrong?

Your generalized Lagrange function:
$$L=x+2y+\lambda_1(1-x^2-y^2)+\lambda_2(-5-3x-4y)$$
The Kuhn-Tucker conditions:
$$\begin{cases}L_x\ge 0, xL_x=0\\
L_y\ge 0,yL_y=0\\
L_{\lambda_1}\le 0,\lambda_1\ge 0,\lambda_1L_{\lambda_1}=0\\
L_{\lambda_2}\le 0,\lambda_2\ge 0,\lambda_2L_{\lambda_2}=0\\
\end{cases} \Rightarrow 
\begin{cases}1-2x\lambda_1-3\lambda_2\ge 0,xL_x=0\\
2-2y\lambda_1-4\lambda_{\color{red}2}\ge 0,yL_y=0\\
x^2+y^2\le 1,\lambda_1\ge 0,\lambda_1L_{\lambda_1}=0\\
3x+4y\le -5,\lambda_2\ge 0,\lambda_2L_{\lambda_2}=0\end{cases}$$
When $\lambda_1>0,\lambda_2>0$ we get:
$$\begin{cases}L_{\lambda_1}=0\\ L_{\lambda_2}=0\end{cases} \Rightarrow \begin{cases}x^2+y^2=1\\ 3x+4y=-5\end{cases} \Rightarrow (x,y)=\left(-\frac35,-\frac45\right).$$
