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So, I have a function $$ f(x, y) = x^2-4xy+4y^2 $$ subject to constraint $$ g(x, y) = x^2+y^2 = 1 $$ The task asks to find the maxima and minima values using Lagrangian.

I found the gradients: $ \begin{equation} \nabla f(x, y) = (2x-4y, -4x+8y) \end{equation} $ and $ \begin{equation} \nabla g(x, y) = (2x, 2y) \end{equation} $

Solving for the system $$ \begin{cases} 2x-4y=2\lambda x \newline -4x+8y=2\lambda y \newline x^2+y^2=1 \end{cases} $$ I decided to take an eigenvalue approach to obtain values for $\lambda$: $$A=\begin{bmatrix} 1-\lambda & -2\\ -1\over2 & 1-\lambda\\ \end{bmatrix}$$ We obtain $\lambda_{1,2}=0,2$. Substituting them into our system we get: $\begin{equation} x=2y \end{equation} $ for the case of $\lambda_1=0$ and $\begin{equation} x=-2y \end{equation} $ for the case of $\begin{equation} \lambda_2=2 \end{equation} $

I have no idea how to proceed from here, since I am not really getting good-looking values of $x$ and $y$. I am quite confused. Any feedback along with correction of mistakes would be highly appreciated.

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By C-S $$x^2-4xy+4y^2=(x-2y)^2\leq(x^2+y^2)(1^2+(-2)^2)=5.$$ The equality occurs for $(x,y)||(1,-2)$, which says that $5$ is a maximal value.

In another hand, $$x^2-4xy+4y^2=(x-2y)^2\geq0,$$ where the equality occurs for $x=2y$.

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Do a trigonometric change of variables to make the problem unconstrained: $x = cos(\theta)$ and $y=sin(\theta)$. Plug it in and find where the gradient vanishes. It will give you both the maximum of the minimum. If my calculations are not wrong the extremum points are given by the equality $tan(\theta) = \frac{4}{3}$.

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Your answer $x=2y$ is correct and gives the minimum value of 0.

The other answer (with $\lambda_2=2$) actually gives $y=-2x$. Then $x=-\frac{1}{\sqrt5}$, $y=\frac{2}{\sqrt5}$ gives the maximum value of 5.

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