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!
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 $\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)$:

$$\begin{align} \dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&amp;=2x+\lambda\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&amp;=2y+\lambda\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&amp;=x+y-1 .\end{align}$$

Next solve the system:

$$\begin{align} \begin{cases}\dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&amp;=0\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&amp;=0\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&amp;=0  \end{cases}&amp;\iff \begin{cases}\\2x+\lambda=0\\
2y+\lambda=0\\
x+y-1=0 \end{cases}\\
&amp;\iff  \begin{cases}\\2x=-\lambda\\
2y-2x=0\\
x+y-1=0 \end{cases}\\
&amp;\iff \begin{cases}\\2x=-\lambda\\
x=y\\
2x=1 \end{cases}\\
&amp;\iff x=\dfrac 1 2=y \land \lambda =-1.\end{align}$$

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.
$$ F(x,y,\lambda) = f + \lambda g = x^2 + y^2 + \lambda(x+y - 1) $$

$$ F_x = 2x + \lambda = 0 $$

$$ F_y = 2y + \lambda = 0 $$

$$ F_\lambda = x + y - 1 = 0 $$

$$ \Rightarrow x = y = \frac{1}{2} $$