<s>[INST] Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$ Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $\bigtriangledown f=\lambda\bigtriangledown g$, so from this I get $(3y,3x)=\lambda(2x+y,2y+x)$ so $\lambda=\frac{3y}{2x+y}$ and $\lambda=\frac{3x}{2y+x}$ so $\frac{3y}{2x+y}=\frac{3x}{2y+x}$ so $x^2=y^2$ so $y=\pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (\sqrt{3},-\sqrt{3})$ and $(-\sqrt{3},\sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
 [/INST] 1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=\pm √3; y=\mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =\pm 1$, $y=\pm 1$.
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