Find the absolute maximum and minimum for $F(x,y)=x^2+3y^2-y$ with the condition $x^2+2y^2 \leq 1$ Find the maximum and minimum of the function :
$$F(x,y)= x^2 +3y^2-y$$ with the condition :
 $$x^2+2y^2 \leq 1$$
Can this question be solved by calculus or by some other way? 
 A: It can be solved by calculus. You need to find the partial derivatives of $F$. You have
$$F(x,y) = x^2+3y^2-y,$$
and
$$\frac {\partial F}{\partial x} = 2x,$$
and
$$\frac {\partial F}{\partial y} = 6y-1.$$
Equate these partial derivatives to $0$ to get $x = 0$ and $y = \frac{1}{6}$. This is a point in the interior of the given domain, as $0^2+2\cdot(\frac{1}{6})^2<1$, so you can check the value of $F$ at this point.
You also need to consider the behavior of $F$ on the boundary $B$ of the given domain, that is, on the set of points $(x,y)$ for which $x^2+2y^2 = 1$. From this equation, I get $x^2 = 1-2y^2$, which I plug into $F(x,y)$:
$$F_B(y) = 1-2y^2+3y^2-y =1+y^2-y.$$
Here I use $F_B$ as the restriction of the function $F$ to the boundary $B$. Notice that $F_B$ is a function of just one variable. Differentiating this, I get
$$F_B'(y) = 2y-1,$$
and equating it to $0$, I get $y = \frac{1}{2}$. From $x^2 = 1-2y^2$, I see that it corresponds to $x = \frac{1}{\sqrt{2}}$, so $(x,y) = (\frac{1}{\sqrt{2}},\frac{1}{2})$ is another point of interest.
So you just need to check the value of $F$ at the two points that we found above.
