Use Lagrange multiplier to find absolute maximum and minimum Use Lagrange multiplier to find absolute maximum and minimum of $f(x,y) =x^2+xy+y^2, x^2+y^2 =8$. 
What i've done so far..  
$f_x = \lambda g_x \Rightarrow 2x+y =\lambda2x, \\f_y = \lambda g_y \Rightarrow x+2y = \lambda 2y,\\g(x,y) = x^2+y^2 -8 =0$ 
May I know how should i proceed from here?
 A: You have to determine $x,y,\lambda$. One possible way: 
$$x = 2y \cdot (\lambda-1)\\
\stackrel{eq. 1}{\Rightarrow} 4y \cdot (\lambda-1) +y = 2 \lambda \cdot 2y \cdot (\lambda-1) \\
\Leftrightarrow y \cdot (8\lambda-3-4\lambda^2)=0$$
There are two cases:


*

*$y=0$: From the first equation follows $x=0$, thus $x^2+y^2=0 \not= 8$. Can't be true...

*$-4\lambda^2+8\lambda-3=0$: Determine $\lambda$. Afterwards solve $x^2+y^2=8$ by using $x = 2y \cdot (\lambda-1)$. 

A: On a side note, this problem can be solved very nicely with substitution since it's equivalent to:
$$f(\theta) = 8 \left ( 1 + \frac{1}{2}\sin 2\theta \right )$$
where $ x = r \cos\theta$, $y = r \sin \theta$ and $r = 2\sqrt{2}$.
Since $f(\theta)$ is the same as maximizing $\sin 2 \theta$, $f(\theta)$ is maximum and minimum at $\theta = \pi n + \frac{\pi}{4}$ and $\theta = \pi n - \frac{\pi}{4}$ respectively.
So if $n = 1$, we have a maximum at $x = -2, y = -2$ and a minimum at $x = -2, y = 2$.
A: The best way (I think) to do this is to use the function 
$$\Lambda (x,y,\lambda)=f(x,y)+\lambda g(x,y)=x^2+y^2+xy+\lambda(x^2+y^2-8)$$
then solve the following system of equation
$$\left\{\begin{matrix}
\cfrac{d\Lambda}{dx}=0&&&&&&(1)\\
\cfrac{d\Lambda}{dy}=0&&&&&&(2)\\
\cfrac{d\Lambda}{d\lambda}=0&&&&&&(3)\\
\end{matrix} \right.$$
Find $x$ and $y$ in terms of $\lambda$, then find $\lambda$ and substitue back to find the values of $x$ and $y$.  
A: $2x+y =\lambda2x, \\x+2y = \lambda 2y $
Once you have these equations, you should see that it is merely a problem of determining the eigenvalues $\lambda_1 , \lambda_2 $ of the matrix $M = \begin{pmatrix}
1 & 1/2 \\
1/2 & 1 \\
\end{pmatrix}$ . One of them will yield you the absolute minimum, and the other the absolute maximum.
This is done by solving the characteristic polynomial. 
$
\left| \begin{array}{ccc}
1-\lambda & 1/2 \\
1/2  & 1-\lambda \\
\end{array} \right| = 0
$
Which yields:
$ (1-\lambda)^2 = 1/4 \rightarrow \lambda_1 =1/2 ,\lambda_2 =3/2$ 
Now you have either of the two cases:
 $2x+ y = x \rightarrow (x,y)=(-1,1)*t\\
  2x+ y = 3x \rightarrow (x,y) = (1,1)*t $
In order to have them with length of 8, you should find the relevant $t$ for which $|| (-t,t)||^2=8, ||(t,t)||^2=8$

Another way to think about your problem is that you have a quadratic form $ ax^2 + 2bxy + cy^2  $, where $ a = 1, b= 1/2 , c = 1$  . The maximal and minimal values of a quadratic form, under the constraint of constant length of $(x,y)$ is attained in its eigenvalues, and $(x,y)$ direction is the direction of the eigenvectors.
