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.
 A: If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not.  That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = \pm 1\\
x = \pm 1\\
y = \pm 1$
$f$ is minimized when $u = 0$
$v = \pm \sqrt 3\\
x = \pm \sqrt 3\\
y = \mp \sqrt 3$
A: Let $3xy=k$.
Thus, 
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $k\neq0$.
Thus, $$(k-9)^2-4k^2\geq0$$ or
$$(k+9)(k-3)\leq0$$ or
$$-9\leq k\leq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-\frac{k-9}{2k}y$, 
which says that
$$\max_{x^2+y^2+xy=3}3xy=3$$ and
$$\min_{x^2+y^2+xy=3}3xy=-9$$
A: If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(\sqrt3y)^2$$
Let $\sqrt3y=\sqrt{12}\cos t\iff y=2\cos t$
$2x+y=2\sqrt3\sin t\iff 2x=2\sqrt3\sin t-2\cos t\iff  x=\sqrt3\sin t-\cos t=2\sin\left(t-30^\circ\right)$
$3xy=12\cos t\sin\left(t-30^\circ\right)=6[\sin(2t-30^\circ)-\sin30^\circ]$
Now for real $x,y;t$ is real $$-1\le\sin(2t-30^\circ)\le1$$
A: Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (\sqrt{3},-\sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ \pm (\sqrt 3, -\sqrt 3)$
A: 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 $x=y$.
$x^2=1$; $x =\pm 1$, $y=\pm 1$.
A: As the equations involved are homogeneous, making $y = \lambda x$ we have
$$
\min(\max)3x^2\lambda\ \ \mbox{s. t. }\ \ x^2(1+\lambda^2+\lambda) = 3
$$
which is equivalent to
$$
\min_{\lambda}(\max_{\lambda})f(\lambda) = \frac{9\lambda}{\lambda^2+\lambda+1}
$$
so the stationary points are at
$$
f'(\lambda) = -\frac{9 \left(\lambda ^2-1\right)}{\left(\lambda ^2+\lambda +1\right)^2} = 0\Rightarrow \lambda^* = \{-1,1\}
$$
then the solutions are at $y = \pm x$ etc.
