Finding the maximum value given two system of equations

Question

I was given $(x,y)$ that satisfies both of this equation:

$4|xy| - y^2 - 2 = 0\\(2x+y)^2 + 4x^2 = 2$

And was asked to find the maximum value of $4x + y$.

Solving for $y^2$, I get this equation: $8x^2 + 4|xy| + 4xy - 4 = 0$

I assumed that if I were to find the maximum value of $4x + y$, then both $x$ and $y$ must be positive, so $xy \ge 0$ means that $|xy|$ is equal to $xy$. So, I have

$8x^2 + 8xy - 4 = 0$

Solving for $y$, I have $y = \frac{4-8x^2}{8x}$ which makes $4x+y = f(x) = \frac{24x^2 + 4}{8x}$. Finding the extreme point using $f'(x) = 0$, I get $x = \frac{1}{\sqrt{6}}$, $y = \frac{\sqrt{6}}{3}$, and $f(\frac{1}{\sqrt{6}}) = \sqrt{6}$, which is the wrong solution (and this being the minimum point, too, while not satisfying the first equation at the same time ...)

The choices were: $1, \sqrt{2}, 2, \sqrt{3}, 4$. Some help would be helpful!

Answer 1

From the two equations you get $$4|xy|-y^2=(2x+y)^2+4x^2 \implies 2|xy|=4x^2+y^2+2xy.$$

If $xy \geq 0$, then we get $$2xy=4x^2+y^2+2xy \implies 4x^2+y^2=0 \implies x=y=0.$$ In which case the expression $\boxed{\color{blue}{4x+y=0}}$.

If $xy < 0$, then we get $$-2xy=4x^2+y^2+2xy \implies 4x^2+y^2+4xy=0 \implies (2x+y)^2=0.$$ Thus $2x+y=0$. Now the expression $\color{blue}{4x+y}=2x+(2x+y)=\boxed{\color{red}{2x}}$.

But with $2x+y=0$, we also get $y=-2x$ and plugging that in one of the given equations we get $x=\pm\frac{1}{\sqrt{2}}$.

So the max. value of the expression $\color{blue}{4x+y}=\color{red}{2x}=\sqrt{2}$.

Answer 2

If $xy\geq0$ so $$x=\frac{y^2+2}{4y},$$ which after substitution to $(2x+y)^2+4x^2=2$ gives $$10x^4+8y^2+8=0,$$ which is impossible.

If $xy<0$, so $$x=-\frac{y^2+2}{4y},$$ which after substitution to $(2x+y)^2+4x^2=2$ again gives $$y^2=2$$ and $$(x,y)=\left(-\frac{1}{\sqrt2},\sqrt2\right)$$ or $$(x,y)=\left(\frac{1}{\sqrt2},-\sqrt2\right)$$ Can you end it now?