Maximize $f(x,y)=xy$ subject to $x^2-yx+y^2 = 1$ 
Use Lagrange multipliers method to find the maximum and minimum values of the function 
  $$f(x,y)=xy$$
  on the curve
  $$x^2-yx+y^2=1$$

Attempt:
First I set let $g(x,y)=x^2-xy+y^2-1$ and set $$\nabla f=\lambda\nabla g$$
so
$$(y,x)=\lambda(2x-y,2y-x)$$
then
$$\begin{cases} 
\lambda=\frac{y}{2x-y} & (1) \\
\lambda=\frac{x}{2y-x} & (2)\\
x^2-yx+y^2=1
\end{cases}
$$
Solving $(1)$ and $(2)$ simultaneously, I get that $$y^2=x^2$$
Substitutiting into $(3)$ and following through with the arithmetic, I get four candidates for max and min, namely $$(1,1),(-1,-1),\big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big),\big(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)$$
 Evaluating these points on $f$, I get that the maximum value is $$1 \ \text{at} \ (\pm1,\pm1)$$
and the minimum value is $$-\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big)$$
Am I correct? I am unsure if there are indeed four critical points.
 A: As a check, not using Lagrange multipliers:
$x^2-xy+y^2 =1$.
1) Minimum.
$(x+y)^2 -3xy =1.$
$3xy= (x+y)^2 -1 ;$
Minimum of $f(x,y) =-(1/3).$
2) Maximum.
$(x-y)^2 +xy =1;$
$xy = 1- (x-y)^2;$
Maximum of $f(x,y) = 1.$
A: Geometrical ways:
\begin{align}
   x^2-xy+y^2 &= 1 \\
  \frac{(x+y)^2}{4}+\frac{3(x-y)^2}{4} &= 1 \tag{1}
\end{align}
With the transformation $(X,Y)=
\left( \dfrac{x+y}{\sqrt{2}}, \dfrac{x-y}{\sqrt{2}} \right)$,
$$\frac{X^2}{2}+\frac{3Y^2}{2}=1 \tag{2} $$
which is an ellipse with semi-major and minor axes $\sqrt{2}$ and $\sqrt{\dfrac{2}{3}}$ respectively.
Also, $$xy=\frac{X^2-Y^2}{2}$$

Now $(2)$ touches
  
  
*
  
*$X^2-Y^2=2$ at $\left( \pm \sqrt{2},0  \right)$ which gives maximum $xy$ of $1$
  
*$X^2-Y^2=-\dfrac{2}{3}$ at $\left( 0, \pm \sqrt{\dfrac{2}{3}} \right)$ which gives minimum $xy$ of $-\dfrac{2}{3}$


A: The lagrange function is $L(x,y,\lambda)=f(x,y)-\lambda g(x,y)$
To take out the maximum of this function you (effectively) partially differentiated both R.H.S and L.H.S and found the the value of lambda for which L is maximum by equating these differentials to zero i.e $ \partial L=0=\partial f-\lambda\partial g$.
This gave you $L(+_- 1,+_-1,1)$
Which would then (being global maximum by definition) be greater than equal to the function itself and hence the inequality will result.Hope this helps.
