I am supposed to find all the critical points $f(x,y,z) = {x{y^2}}$ when subjected to $x+2y=3$ and $x-y=0$.
I used Lagrange multiplier method and i get x = y = 1, by solving the 2 constraint equations. Are there any other critical points ? How can i find them ?
How can i classify this point (1,1) ? Is it maxima or minima.
If the problem is to find the critical points of $f(x,y)=e^{xy^2}$ on the region $$M = \{(x,y)\in\mathbb{R}^2:x+2y=3,x-y=0\}$$ notice that $M=\{(1,1)\}$ is only one point so $f\vert_M$ is constant on that region and you don't need Lagrange multipliers for that as every point of $M$ is a maximum and a minimum of $f\vert_M$.
However, if the problem is to find the maxima and minima of $f$ on $$M_1 = \{(x,y)\in\mathbb{R}^2:x+2y=3\} \text{ and } M_2 = \{(x,y)\in\mathbb{R}^2:x-y=0\}$$ separately, note that, on $M_1$, you have $x=3-2y$ so $f\vert_{M_1}(x,y)=e^{(3-2y)y^2}=h(y)$ which is a function whose maxima and minima can be easily found by equating $h'=0$: maximum at $y=1$ and minimum at $y=0$ so $f\vert_{M_1}$ has a maximum at $(1,1)$ and a minimum at $(3,0)$. The same goes with $f\vert_{M_2}$ as you have that $x=y$ which means that $f\vert_{M_2}(x,y)=e^{x^3}=g(x)$. However, since $g$ is strictly increasing, $f$ won't have neither a maximum nor a minimum on $M_2$. This is totally fine as $M_2$, while it is closed, it isn't bounded and, thus, it isn't compact either.