Extremum values using Lagrange multiplier

Question

Given $f(x,y) = 2x^2-8xy+9y^2$ on $x^2+y^2=1$

now I am stuck in solving equations

$2x - 4y + \lambda x = 0$

$-4x + 9y + y \lambda = 0$

how do I do ?

Answer 1

$f(x,y) = \begin{bmatrix} x&y\end{bmatrix} \begin{bmatrix} 2&-4\\-4&9\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$

The constraint tells us that $\|(x,y)\| = 1$

$f(x,y)$ will be maximized when $(x,y)$ fall on the largest eigenvector, and minimized when it falls on the smallest.

$\lambda^2 - 11\lambda + 2 = 0\\ \lambda = \frac {11 \pm \sqrt{113}}{8}$

Now, if you still want to use Lagrange multipliers...

$\lambda = \frac{4y-2x}{x}\\ \lambda = \frac{4x-9y}{y}\\ \frac{4x-9y}{y} =\frac{4y-2x}{x}\\ 4x^2 - 7xy - 4y^2 = 0\\ x^2 + y^2 = 1\\ 4 - 7xy - 8y^2 = 0\\ x = \frac {4-8y^2}{7y}\\ (\frac{4-8y^2}{7y})^2 +y^2 = 1\\ \frac{16-64y^2 + 64y^4}{49y^2} + y^2 = 1\\ 113y^4 -113y^2 + 16 = 0\\ y^2 = \frac 12 \pm \frac{7}{2\sqrt{113}}$

$y = \sqrt{\frac 12 + \frac{7}{2\sqrt{113}}}, \sqrt{\frac 12 - \frac{7}{2\sqrt{113}}}$

And I am getting a little bit exhausted. You can take it from here.

Answer 2

Let $L(x,y,\lambda)=2x^2-8xy+9y^2+\lambda(x^2+y^2-1).$

Thus, in the critical point we obtain: $$\frac{\partial L}{\partial x}=0$$ and $$\frac{\partial L}{\partial y}=0$$ or $$4x-8y+2\lambda x=0$$ and $$18y-8x+2\lambda y=0$$ or $$2xy-4y^2+\lambda xy=0$$ and $$9xy-4x^2+\lambda xy=0,$$ which gives $$2xy-4y^2=9xy-4x^2$$ or $$4x^2-7xy-4y^2=0.$$ Can you end it now?