I am trying to solve this simple optimization problem:
Minimize $f(x,y)=x^2+y^2$ over $\frac{1}{4}x^2+\frac{1}{9}y^2-1=0$
I get
$L = x^2+y^2+\alpha(\frac{1}{4}x^2+\frac{1}{9}y^2-1)$
then
$0=\frac{\partial L}{\partial x} = 2x+\frac{1}{2}\alpha x$ $\rightarrow$ x=0
$0=\frac{\partial L}{\partial y} = 2y+\frac{2}{9}\alpha y$ $\rightarrow$ y=0
So we get $L=-\alpha$ and $0=\frac{\partial L}{\partial \alpha}=-1$
Anything to help me understand this is greatly appreciated. Thanks!
You have an extremum of $f(x,y)$ on the curve when the derivative is zero. An extremum is either a maximum or a minimum, not necessarily a minimum.
In this particular problem, there are four extrema. It so happens that two of them occur when $x = 0$ and the other two occur when $y = 0$.
There is no extremum on the curve where $x = y = 0$ simultaneously; in fact, there is no point on the curve at all where that condition occurs. Therefore, while setting $x = y = 0$ satisfies both of the equations $\frac{\partial L}{\partial x} = 0$ and $\frac{\partial L}{\partial y} = 0$, it is not a solution. You need to look for other ways to satisfy those equations, for example by the choice of $\alpha$.