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Find the points closest and farthest from 0 in $\mathbb{R}^2$ under the constraint $\left=70$ where $$A=\begin{pmatrix} 3&-2 \\ -2&6 \end{pmatrix}. $$

My work:

So I rephrased the question as: find the min and max points of $f(x,y)=x^2+y^2$ under the constraint $g(x,y)=3x^2-4yx+6y^2-70=0$.

So by Lagrange theorem we will get these 2 equations:

$$2x=\lambda(6x-4y)$$

$$2y=\lambda(12y-4x)$$

I tried to divide the 2 equations but in the end I got to this equation: $$2x^2-3xy-2y^2=0. $$ And now I'm stuck trying to find $x$ or $y$. Can anyone help?

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    You have to use the equation for the constraint – 2017-01-28

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It is actually better to multiply by $\lambda$ on the other side of the equation to avoid fractions. Then you get the system of equations:

$$ 2 \lambda x = (6x - 4y) \iff (\lambda - 3)x + 2y = 0, \\ 2\lambda y = (12y - 4x) \iff 2x + (\lambda - 6)y = 0. $$

For a fixed $\lambda$, this is a system of two linear homogeneous equations in $x,y$. In order to have a non-trivial solution, we must have

$$ \det \begin{pmatrix} \lambda - 3 & 2 \\ 2 & \lambda - 6 \end{pmatrix} = (\lambda - 3)(\lambda - 6) - 4 = 0. $$

The solutions are $\lambda = 2$ and $\lambda = 7$. Now you can solve the linear system and use the constrain to find the maximum and the minimum values.

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    Thank you for answering! Excuse me for my amateur linear algebra but why exactly must we have $\det \begin{pmatrix} \lambda - 3 & 2 \\ 2 & \lambda - 6 \end{pmatrix} = 0$? Can you elaborate or link me to something that explains it? – 2017-01-28
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    A homogeneous system $Ax = 0$ has a non-trivial solution iff $A$ is not invertible iff $\det(A) = 0$. – 2017-01-28
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    (and you are looking for a non-trivial solution because obviously $(x,y) = (0,0)$ doesn't satisfy the constraint). – 2017-01-28