Use Lagrange multiply to find the local extremes of $f$

Question

Use Lagrange multiply to find the local extremes of $f$

Let $f(x,y)=y^2-4xy+4x^2$ and $x^2+y^2=1$

My work: Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $f(x,y)=y^2-4xy+4x^2$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $g(x,y)=x^2+y^2-1$

We need define the function of Lagrange $$L(x,y,\alpha)=f(x,y)+\alpha(g(x,y))$$

Then, calculating the critic point of $L$ we have

$\begin{equation} -4y+8x+2x\alpha=0\\ 2y-4x+2\alpha y=0\\ x^2+y^2-1=0 \end{equation}$

Then, solving the system, we have $x=0,\, \, y=0$

But i have a problem with this. this answer dont have sense because if $x=0$ and $y=0$ then $-1=0$

Can someone help me?

Answer 1

If you solve the first two equations in your system of three equations for $\alpha$ you get

$$ \alpha=-4+2\frac{y}{x}=-1+2\frac{x}{y} $$

Substituting $u=\dfrac{y}{x}$ into $$ -4+2\frac{y}{x}=-1+2\frac{x}{y} $$

gives the equation

$$ 2u^2-3u-2=0 $$

with solutions $u=-\dfrac{1}{2}$ or $u=2$ leading to the result that $y=-\frac{1}{2}x$ or $y=2x$.

Substitute these into the third equation in your system of three equations and you will find eight solutions for $(x,y)$. Evaluate $f(x,y)=y^2-4xy+4x^2=(y-2x)^2$ at each of these eight solutions and you will find your extrema.

Answer 2

The equations become $\begin{bmatrix} 4 & -2 \\ -2 & 1\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} + \alpha \begin{bmatrix} x \\ y\end{bmatrix} = 0$.

You are looking for unit vectors that are eigenvectors of the above.