Supremum and infimum of function of two variables Consider $D = \left \{ x \in \mathbb{R} : x_1^2 + 44x_2^2 \leqslant 5 \right \}$ and function $f: D \rightarrow \mathbb{R}$, $f(x) = 13x_1 - 22x_2$.
Find supremum and infimum of $f$. For both of them examine if they are reached inside $D$.
To find sup and inf I calculated partial derivatives, they are both non-negative, so the function doesn't have any local minimums/maximums.
So I need to examine value of function on the edge - $\left \{ x \in \mathbb{R} : x_1^2 + 44x_2^2= 5 \right \}$.
How to do that?
 A: As I mentioned in my comment $f_{x_1} = 13$ and $f_{x_2}= -22$, and so this function has no critical points. Because $D$ is closed and bounded $f$ achieves its suprema and infima in $D$ and so we can search directly for the max and the min. Because there are no critical points we know this has to occur on the boundary.
There are two ways the search on the boundary, the first is using Lagrange multipliers, the second is uses the constraint $x_1^2 + 44x^2_2 = 5$ to reduce $f$ to (two) functions of one variable, and then use single-variable calculus methods. I'll show you the first.
Let $G(x_1,x_2) = x^2_1 + 44x^2_2 -5$. Then we know that the extrema of $f$ subject to $G$ occur when $\lambda \nabla f =  \nabla G$ for some constant $\lambda$. In particular
$$
13 \lambda = 2x_1,\qquad -22\lambda = 88x_2
$$ 
And so, from the first equation $\lambda = \frac{2}{13}x_1$, and plugging this into the second $-\frac{2}{13}x_1 = 4x_2$, or $x_1 = -26x_2$. You can use this with the fact that $x_1^2 + 44x_2^2 = 5$ to solve explicitly for the points on the boundary where $f$ has its extrema.
