1
$\begingroup$

Let $f(x,y)=e^{x^2-xy+y^2}$

(a) Find all the critical points of the following function.

(b) Find the all the local maxima and local minima of the function if there is any.

What i tried.

I tried to differentiate the following function wrt $x$ and $y$, and then equating them to $0$ ie $$(2x-y)e^{x^2-xy+y^2}=0$$ and $$(2y-x)e^{x^2-xy+y^2}=0$$

While i could differentiate the function, the diffculty lies in solving the following equations to get $x$ and $y$. I only managed to get a critical point of $(0,0)$ but im sure there are other critical points as well .Could anyone explain.Thanks

$\endgroup$
2
  • $\begingroup$ No, there aren't other critical points. What you did is correct. $\endgroup$
    – Git Gud
    Commented Nov 20, 2014 at 16:01
  • 1
    $\begingroup$ The exponential cannot be $0$. Then, what is left ? Exactly what you got ! Cheers :-) $\endgroup$
    – Claude Leibovici
    Commented Nov 20, 2014 at 16:02

1 Answer 1

Reset to default
1
$\begingroup$

Notice that $e^{x^2-xy+y^2} > 0$ for every pair $(x,y)$. Then what you found boils down to $$\begin{cases} 2x - y = 0 \\ 2y - x = 0 \end{cases} ,$$ which has as the only solution $(0,0)$. For the second part, look at ${\rm Hess} \ f_{(0,0)}$, and see if is is positive/negative definite, etc.

$\endgroup$
1
  • $\begingroup$ It seems correct to me. You can think that $x^2-xy + y^2 \geq 0$ for all $(x,y)$, and $t \mapsto e^t$ is a crescent function. So $e^0 = 1$ is the minimum value.. $\endgroup$
    – Ivo Terek
    Commented Nov 20, 2014 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .