0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.