Find extremal values of $f(x,y)=x^2+4y^2$ on the set $D$. Find abs.min and max values of $f(x,y)=x^2+4y^2$ on the set $D=\{(x,y) \in \mathbb{R}^2; x^2 + (y+1)^2 \leq 4, y \geq -1, y \leq x+1\}$. 
For the interior, it is as usually $f_x=0$ and $f_y=0$, and I got $(0,0)$ as a critical point, and $f(0,0) =0$.
For the boundary, we go along $y=x+1$ and $x \in [-2,0]$. Then $f(x,x+1)=5x^2+8x+4$ and we can see $f'(x)=10x+8$ where we get $10x+8=0 \iff x= \frac{-4}{5}$.  Evaluating $f$ at the endpoints and critical points we get rel.min and max at points $$f(\frac{-4}{5},\frac{1}{5})=\frac{4}{5}, f(-2,-1)=8,f(0,1)=4.$$
Similarly we go along $y=-1$ and $x \in [-2,2]$, where $f(x,-1)=x^2+4$ and $f'(x)=2x$ and so $f'(x)=0 \iff x=0$. Evaluating $f$ we get $$f(-2,-1)=8,f(2,-1)=8,f(0,-1)=4.$$
Finally we go along $y=\sqrt{x^2-4}-1$ where $x \in[0,2]$. Then in a similar fashion $f(x,\sqrt{x^2-4}-1)=5x^2-12-8\sqrt{x^2-4}$ and its derivative equals zero iff $x=\pm \sqrt \frac{116}{25}$ and this is approximately 2.15 which will not be in the interval $[0,2]$. Thus maximum value is $8$ while the minimum is $0$. Can someone let me know if this approach is correct, or if I should use some other method?  
Update: in the last step, we go along $y=\sqrt{4-x^2}-1$ where $x \in[0,2]$. Then in a similar fashion $f(x,\sqrt{4-x^2}-1)=5x^2+20-8\sqrt{x^2-4}$ and its derivative equals zero iff $x=\pm \sqrt \frac{84}{25}$. Now since $x \in[0,2]$ the value $x=- \frac{\sqrt{84}}{5}$ is not in the interval so we have only check $f(\frac{\sqrt{84}}{5},-\frac{1}{5})$.

Your approach seems to be correct. However, note that you are able to in fact "talk" about extremum values since the function $f(x,y)$ is continuous on a compact set $D$.
