<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] Those called homogeneous problems can be easily solved 
Calling $d^2 = x^2+y^2$ and making $y = \lambda x$ we have
$$
d^2 = x^2(1+\lambda^2)
$$
but from $3x^2+4 x y +6 y^2 = 140$ we obtain
$$
x^2(3+4\lambda+6\lambda^2) = 140
$$
then
$$
d^2 = \frac{140(1+\lambda^2)}{3+4\lambda+6\lambda^2}
$$
now deriving
$$
\frac{d}{d\lambda}d^2 = \frac{200(2\lambda^2-3\lambda-2)}{(3+4\lambda+6\lambda^2)^2}
$$
and the condition for local minimum/maximum is
$$
2\lambda^2-3\lambda-2 = 0
$$
obtaining $\lambda = \{-\frac 12,  2\}$ and
$$
\min d^2 = 20\Rightarrow d = \sqrt{20}\\
\max d^2 = 70\Rightarrow d = \sqrt{70}
$$
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