Applying the Lagrangian function to find critical points So I have the following function
$$ f(x,y) = x^2+y^2 $$
subject to 
$$ g(x,y) = x+y-1 = 0. $$
And I have to use the Lagrangian to find the critical points, and determine wether they are constrained minima, constrained maxima or neither.
I am starting on this guy right now. I have no clue, and any hints would be great (not necessarily an answer). Im on wikipedia, which is making me more confused right now!
 A: For your better reading, I'll follow the notation used on wikipedia.
Let $(x,y)\in \Bbb R^2$.
Start of by defining $\Lambda(x,y,\lambda):=f(x,y)+\lambda g(x,y)$.
Then find $\dfrac{\partial\Lambda }{\partial x}(x,y,\lambda), \dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)$ and $\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)$:
$$\begin{align} \dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&=2x+\lambda\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&=2y+\lambda\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&=x+y-1 .\end{align}$$
Next solve the system:
$$\begin{align} \begin{cases}\dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&=0\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&=0\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&=0  \end{cases}&\iff \begin{cases}\\2x+\lambda=0\\
2y+\lambda=0\\
x+y-1=0 \end{cases}\\
&\iff  \begin{cases}\\2x=-\lambda\\
2y-2x=0\\
x+y-1=0 \end{cases}\\
&\iff \begin{cases}\\2x=-\lambda\\
x=y\\
2x=1 \end{cases}\\
&\iff x=\dfrac 1 2=y \land \lambda =-1.\end{align}$$
Thus finding the set of critical points: $\left\{(x,y)\in \Bbb R^2\colon x=\dfrac 1 2 = y\right\}=\left\{\left(\dfrac 1 2, \dfrac 1 2\right)\right\}$.
From here I'll let you make the final conclusions.
In order to verify the conclusions you find, note that $x+y-1=0\iff y=1-x$, so you can just replace $y$ with $1-x$ in $f$ and you have reduced the problem to a single variable which you should know how to solve.
A: $$ F(x,y,\lambda) = f + \lambda g = x^2 + y^2 + \lambda(x+y - 1) $$
$$ F_x = 2x + \lambda = 0 $$
$$ F_y = 2y + \lambda = 0 $$
$$ F_\lambda = x + y - 1 = 0 $$
$$ \Rightarrow x = y = \frac{1}{2} $$
