How to solve a system of equation with Lagrange multiplier?

Question

How can I go ahead solving this equation ?

1. $2x = \lambda(10x + 6y)$

2. $2y = \lambda(10y + 6x)$

3. $5x^2 + 6xy + 5y^2 = 1$

My attempt was dividing number $1$ by number two. Eliminating the Lagrange multiplier, and then solving for $x$ and $y$. But that did not work. Do you have any advice or any simpler methods for solving this system of equation?

Answer 1

I would usually solve for $\lambda$ and equate the two values of $\lambda$ which gives $$\frac{2x}{10x+6y}=\frac{2y}{10y+6x}(=\lambda)$$ $$x(5y+3x)=y(5x+3y)$$ $$5xy+3x^2=5xy+3y^2$$ $$x^2=y^2$$ $$y=\pm x$$ Hence the last equation is $$10x^2\pm6x^2=1$$ $$4x^2,16x^2=1$$ $$x^2=\frac14,\frac1{16}$$ $$x=\pm\frac12,\pm\frac14$$ Hence solutions are $$(\frac12,\frac12),(\frac12,-\frac12),(\frac14,\frac14),(\frac14,-\frac14),(-\frac12,\frac12),(-\frac12,-\frac12),(-\frac14,\frac14),(-\frac14,-\frac14),$$

Answer 2

We have $$\frac{x}{y}=\frac{5x+3y}{5y+3x}$$ or $$x^2=y^2,$$ which gives $x=y$ or $x=-y.$

Can you end it now?

The original problem you can solve without LM: $$2(x^2+y^2)\leq5x^2+6xy+5y^2\leq8(x^2+y^2),$$ which gives $$\frac{1}{2\sqrt2}\leq\sqrt{x^2+y^2}\leq\frac{1}{\sqrt2}.$$