I have the function $f(x,y)=e^{-x-2y}x^3y^4$.

I know that at the stationary values $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$.

Upon solving the equations I find that I get the following:

$x^2(x-3)=0$ and $y^3(y-2).$

This gives me x=0,3 and y=0,2 . So my conclusion was that there are stationary points all along the lines x =0 x=3 y=0 and y=2. Plotting it I found the entire lines x =0 and y =0 were stationary, but not along x =3 and y =2, only the point (3,2) was stationary.

How can I tell when I have a stationary line as opposed to a stationary point?


Note that when $x = 0$, $f(0,y) = 0$ for all $y$; and when $y = 0$, $f(x,0) = 0$ for all $x$. So both $\partial f/\partial x$ and $\partial f/\partial y$ vanish whenever $x = 0$ for any $y$, or when $y = 0$ for any $x$. That's why the lines $x = 0$ and $y = 0$ are stationary, but the nontrivial $(x,y) = (3,2)$ is an isolated stationary point. Note that in this case, in a neighborhood of $(3,2)$, the function has a local maximum: both $\partial^2 f/\partial x^2$ and $\partial^2 f/\partial y^2$ are negative.


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