# Query about finding stationary points of a 2 variable function.

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.