Find extreme values of the function $f(x,y) = xy$ on the surface $g(x,y) = \frac {x^2}{8} + \frac {y^2}{2} - 1 = 0$.
My approach :
First I created an auxilliary function $F(x,y,λ) = xy + λ(\frac {x^2}{8} + \frac {y^2}{2} - 1)$ Then I equated $\partial{F/\partial{x}} = 0$, $\partial{F/\partial{y}} = 0$ and $\partial{F/\partial{λ}} = 0$.
I got the stationary points as: $(2,1), (-2,1), (-2,-1), (2,-1)$, where $λ = \frac{-x}{y}$.
How to proceed from here ? Should I directly put the stationary points in $f(x,y)$ and get its value ? In that case how will I know which points are extreme as some points may be saddle points also.
I needed a method of solving this question by finding double partial derivative.
Solving $F(x,y,λ) = xy + λ(\frac {x^2}{8} + \frac {y^2}{2} - 1)\Rightarrow \cases{\partial{F/\partial{x}} = 0\\ \partial{F/\partial{y}} = 0\\ \partial{F/\partial{λ}} = 0}$ we obtain four solutions
$$ \left[ \begin{array}{cccc} x y & x & y & \lambda\\ 2 & -2 & -1 & -2 \\ -2 & -2 & 1 & 2 \\ -2 & 2 & -1 & 2 \\ 2 & 2 & 1 & -2 \\ \end{array} \right] $$
as can be depicted in the attached plot. See the four tangent points between the level curves from $x y$ and the elliptical constraint $\frac {x^2}{8} + \frac {y^2}{2} - 1=0$. We have two minima and two maxima.