<s>[INST] Maxima/Minima Problem I am working on this Maxima and Minima Problem :
Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$
I tried it solving using the lagrange's method of multipliers. I get the following equations
$x+3x\theta+2y\theta=0$
$y+4y\theta+2x\theta=0$
$2z=0$
$3x^2+4xy+6y^2=140$
So i get $z=0$
Please suggest how to go about solving for $x$ and $y$ ?
 [/INST] An approach with a bit of geometry:
$3x^2 + 4xy + 6y^2 = 140$ is an ellipse rotated about the origin.
Rewriting 
$3x^2 + 4xy + 6y^2 = 140$;
$2x^2 + (x+2y)^2 + 2y^2 = 140$;
$(x^2 + y^2) = 140/2 - [(x + 2y)^2]/2$.
Let $r$ be the distance from the origin, 
$r^2  = (x^2 + y^2) = 70 - $
$[(x + 2y)^2 ]/2$.
$Max:$
Since the bracket on the r.h.s., a square, is  $\\  \geq 0$, we get for the maximum:
$r^2 = 70$,  at  $ 2y + x = 0$.
The major axis of the ellipse, call it $a$, lies along the line $2y + x = 0$, and  has squared length:
$a^2 = 70$. 
$Minimum:$
The major axis lies along $2y +x = 0$, or  $y = - (1/2)x$ , a straight line with slope $- 1/2$. 
The line $y = 2x$ is perpendicular to $y = - (1/2) x$, the orientation of the major axis, I.e. a line along the minor axis.
Intersection of $y = 2x$ with the ellipse:
$3x^2 + 4x(2x) + 6(2x)^2 = 140$;
$35x^2 = 140$;
$\\x^2 = 4$, and using $y= 2x$, we find 
$y^2 =  4x^2 = 16$.
$Finally$:
$r^2 = min ( x^2 +y^2) = 20$;
For the squared minor axis, call it $b$, we have
$b^2 = 20$.
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