Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$ 
Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$. 

I know how to find maxima/minima if the constraint was the boundary of the circle (by Lagrange's multiplier method). How do I find it for a region? Can I maximise/minimise $f(x,y)$ by finding its saddle points, and then using only those points that lie within the region?

Rewrite constraint as: $\dfrac{x^2}{2} + y^2 \le 1\implies x =r \sqrt{2}\cos \theta, y = r\sin \theta, 0 \le r \le 1, 0 \le \theta \le 2\pi\implies f(x,y) = f(r,\theta) = 2r^2\cos^2 \theta+3r^2\sin^2 \theta- r\sin \theta\implies f_r = 0 = f_{\theta}\implies 4r\cos^2\theta+6r\sin^2\theta-\sin \theta = 0 = -4r^2\sin\theta\cos \theta+6r^2\sin \theta\cos \theta-r\cos \theta\implies 2r^2\sin \theta\cos \theta - r\cos \theta = 0\implies r\cos \theta(2r\sin \theta-1) = 0$. If $r = 0 \implies x = y = 0 \implies f(0,0) = 0$, if $\cos \theta = 0 \implies \sin \theta = \pm 1\implies r = 1/6\implies x = 0, y = 1/6\implies f(0,1/6) = -1/12$, if $2r\sin \theta - 1 = 0 \implies \sin \theta = \dfrac{1}{2r}\implies \sin^2 \theta = \dfrac{1}{4r^2}\implies \cos^2 \theta = \dfrac{4r^2-1}{4r^2}\implies \dfrac{4r^2-1}{r}+\dfrac{3}{2r}-\dfrac{1}{2r}=0\implies \text{no solution this case}$ for $r \ne 0$. Now calculate the values of $f(r,\theta)$ on the boundaries of the rectangle $A_{\theta, r}= \{(r,\theta): 0 \le r \le 1, 0 \le \theta \le 2\pi\}$:
a) $\theta = 0, 2\pi \implies x = r\sqrt{2}, y = 0\implies f(x,y) = 2r^2\le 2 = f(\sqrt{2},0)$
b) $r = 0\implies x= y = 0 \implies f(0,0) = 0$
c) $r = 1 \implies f(x,y) = f(\theta) = 2\cos^2 \theta + 3\sin^2\theta - \sin \theta = \sin^2\theta -\sin \theta + 2 \le \sin^2\theta + |\sin \theta| + 2 \le 1+1+2 = 4$, which attains a value of $4$ when $\theta = \dfrac{3\pi}{2}$.
Thus to sum it up, $f_{\text{min}} = -1/12, f_{\text{max}} = 4$.
