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?
 A: From the point of view of a calculus teacher the first mistake you made was not drawing a picture. Take a look:

To satisfy both the constraints the point $(x,y)$ must be below that line as well as inside the circle. From the picture it looks like the line is actually a tangent to the circle (leaving the proof of that as an exercise for you). The point of tangency, $(-3/5,-4/5)$, is thus the only point satisfying both inequalities, so it is the answer irrespective of the objective function.
When viewing this as a calculus exercise
you need those $\lambda$s only when looking for points that satisfy the constraint with equality.
This is pons asinorum to a general approach (still calculus based) for problems of this type. To actually have an optimization problem I will modify the second constraint to $3x+4y\le0$, meaning that the region of interest is the following.

If you face the task of finding the extrema of a continuosly differentiable function $f(x,y)$ in the shaded region you must consider the following points as candidates:

*

*Any eventual critical points of $f$ in that region. These you find by solving for the points where $\nabla f(x,y)=(0,0)$. Any possible extremum in the interior will be on that list.

*But the extremum may be achieved on the border of the region. You see that the border consists of an arc of a circle and a line segment.

*Should an extremum be on the circle, you can find it by solving the constrained optimization problem using a single Lagrange multiplier for the gradient of $x^2+y^2-1$, the gradient $\nabla f(x,y)$, and the constraint equation $x^2+y^2=1$ to locate these. If a candidate point you find this way is not on the border of the shaded region, throw it out of reckoning.

*Should an extremum be on the line segment,  you can find it by solving the constrained optimization problem using a single Lagrange multiplier for the gradient of $3x+4y$, the gradient $\nabla f(x,y)$, and the constraint equation $3x+4y=0$ to locate these. If a candidate point you find this way is not on the border of the shaded region, throw it out of reckoning.

*Because the shaded region does not have either all of the circle or all of the line in it, it is also possible that extrema you are looking for may fall on the points of intersection of the circle and of the line. These points of intersection are not necessarily solutions to either of the constrained problems in the previous two bullets. Well, we are talking about finitely many points, so just solve for the intersections, and add them to the list of candidates.

*After all this you simply check the values of the objective function at all the candidate points, and pick the maxima/minima.



You have to begin by drawing a picture (when in a low enough dimension). Great for learning.
Only use a Lagrange multiplier, when a constraint is known to be satisfied with equality, when looking for a candidate point of a specific type.
If you need to solve serious problems of this type do learn about the technique described in Farruhota's answer. My first impression is that Karush-Kuhn-Tucker seeks to automate the above process of splitting the border into several components (obviously an advantage in higher dimensions).

A: 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:
$$\begin{cases}L_{\lambda_1}=0\\ L_{\lambda_2}=0\end{cases} \Rightarrow \begin{cases}x^2+y^2=1\\ 3x+4y=-5\end{cases} \Rightarrow (x,y)=\left(-\frac35,-\frac45\right).$$
