Find the maximum and minimum of the function $f$ Find the maximum and minimum of $f(x, y)=xy-y+x-1$ at the set $x^2+y^2\leq 2$. 
I have done the following: 
Since the region $x^2+y^2\leq 2$ is closed, $f$ has a maximum and a minimum,  which is either at the boundary or at the critical points of the function. 
To find the critical points we do the following: 
$$\nabla f=\overrightarrow{0} \Leftrightarrow (y+1, x-1)=(0, 0) \Rightarrow x=1, y=-1$$ 
So, we have that the only critical point is $(1, -1)$, which satisfies the contraint. 
To find the extremas of $f$ at the boundary $x^2+y^2=2$, we will use Lagrange multipliers therem. 
The constraint is $g(x, y)=x^2+y^2-2=0$. 
We are looking for $x$, $y$ and $\lambda$ such that $$\nabla f(x, y)=\lambda \nabla g(x, y) \tag 1$$ and $$g(x, y)=0  \tag 2$$ 
$$(1) \Rightarrow (y+1, x-1)=\lambda (2x, 2y) \Rightarrow \lambda =\frac{1}{2}$$ 
$$\Rightarrow x=y+1$$ 
$$(2) \Rightarrow y=\frac{-1+\sqrt{3}}{2}, y=\frac{-1-\sqrt{3}}{2}$$ 
SO, the extremas are the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$. 
Calculating the values of $f$, we have: $$f\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )=\frac{1}{2} \\ f\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )=\frac{1}{2} \\ f(1, -1)=0$$ 
SO, $f$ has a local maximum at the points $\left (\frac{1+\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2}\right )$ and $\left (\frac{1-\sqrt{3}}{2}), \frac{-1-\sqrt{3}}{2}\right )$ and an absolute minimum at $(1, -1)$. 
Is this correct?? Could I improve something?? 

Looks good, I have just a few comments:
"So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema.
"$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum and minimum"
