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?
 A: For $x=0$ and $y=\frac{1}{6}$ we get ba value $-\frac{1}{12}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^2+3y^2-y\geq-\frac{1}{12}$$ or
$$12x^2+(6y-1)^2\geq0,$$
which is obvious. 
For $x=0$ and $y=-1$ we get a value $4$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$x^2+3y^2-y\leq4$$ or
$$4-3y^2-x^2+y\geq0$$ or
$$16-12y^2-4x^2+4y\geq0$$ or
$$14-14y^2-4x^2+2(y+1)^2\geq0$$ or
$$7(2-2y^2-x^2)+3x^2+2(y+1)^2\geq0,$$
which is obvious.
Done!
A: 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, $f_{\text{min}} = -1/12, f_{\text{max}} = 4$.
A: You can also solve this by using Lagrangian multipliers. You have the problem
$$
\underset{x,y}{\text{minimize}}\ \quad \quad x^2+3y^2-y\\
\text{subject to}\ \ x^2+2y^2-2\leq 0
$$
To put it in Lagrangian form, we simply restate the problem as
$$
L(x,y,\lambda) = x^2+3y^2-y + \lambda(x^2+2y^2-2)
$$
What we need to do now is to find all the KKT points. To do so, we solve the following problem
$$
\frac{\partial L}{\partial x}=0 \iff 2x+2\lambda x=0\quad (1)\\
\frac{\partial L}{\partial y}=0 \iff 6y-1+4\lambda y=0\quad (2)\\
\lambda(x^2+2y^2-2)=0 \quad (3)
$$
First, assume that $\lambda=0$, then we must have (from $(1)$ and $(2)$) $x=0,y=1/6$. This is our first KKT-point. To find the others, assume that $\lambda\neq 0$. Then from $(1)$ we get either $x=0$ or $\lambda=-1$ from $(2)$. This yields the points $(x,y)=(0,1)$, $(x,y)=(0,-1)$, and $(x,y) = (\frac{\pm\sqrt(3)}{2},\frac{1}{2})$. Now lets check which point is minimium/maximum:
$$
f(0,1/6) = -\frac{1}{12}\\
f(0,1) = 2\\
f(0,-1) = 4\\
f(\frac{\sqrt(3)}{2},\frac{1}{2})=1\\
f(\frac{-\sqrt(3)}{2},\frac{1}{2})=1
$$
Thus the minimum is $-\frac{1}{12}$ attained at $(x,y) = (0,\frac{1}{6})$ and the maximum is $4$ attained at $(x,y) = (0,-1)$.
Note: If you had more constraints (equality or inequality) you could use the same approach, however then you would have more Lagrange multipliers. (3) is called complementary slackness. 
A: The global maximum (resp., minimum) of $f$ on the compact elliptical disc $E$ is either taken in the interior of $E$ or on $\partial E$. In the first case the relevant point is a zero of $\nabla f$, in the second case it is a local maximum of the pullback
$$\hat f(\phi):=f\bigl(\sqrt{2}\cos\phi,\sin\phi)=2\cos^2\phi+3\sin^2\phi-\sin\phi=\sin^2\phi-\sin\phi+2\ .$$
The equation $\nabla f(x,y)=(0,0)$ is solved by the point $\bigl(0,{1\over6}\bigr)\in E$, and
$$\hat f'(\phi)=(2\sin\phi-1)\cos\phi$$
has four zeros ${\rm mod}\ 2\pi$, namely ${\pi\over6}$, ${5\pi\over 6}$, ${\pi\over2}$, ${3\pi\over2}$. We therefore obtain a candidate list $L$ consisting of five points:
$$L=\left\{\left(0,{1\over6}\right), \left(\sqrt{3\over2},{1\over2}\right),\left(-\sqrt{3\over2},{1\over2}\right),(0,1),(0,-1)\right\}\ .$$
Computing the value of $f$ at each point $(x_k,y_k)\in L$ will reveal the global maximum and minimum of $f$ on $E$.
