Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and replace $y$ in the quadratic equation with $2x-t$ to get a function but then, I get confused. 
 A: The trick is to notice that:
$$4(2x-y)^2 = 9(2x^2-4xy+6y^2) - 2(x-5y)^2\leq 9(2x^2-4xy+6y^2) = 81$$
It follows that
$$-\frac{9}{2} \leq 2x-y \leq \frac{9}{2}$$
The minimum is attained when $(x,y)=\left(-\dfrac{5}{2},-\dfrac{1}{2}\right)$ and the maximum when $(x,y)=\left(\dfrac{5}{2},\dfrac{1}{2}\right)$.
A: $$2x^2-4xy+6y^2=9$$ and let $$k=2x-y\Rightarrow y=2x-k$$
$2x^2-4x(2x-k)+6(2x-k)^2=9$
$2x^2-8x^2+4xk+6(4x^2+k^2-4xk)=9$
$-6x^2+4xk+24x^2+6k^2-24xk-9=0$
$$18x^2-20kx+6k^2-9=0$$
For real roots, Discriminant $\geq 0$
$$400k^2-4\cdot 18(6k^2-9)\geq 0$$
$$100k^2-108k^2+162\geq 0$$
$$-8k^2+162\geq 0\Rightarrow k^2-\frac{81}{4}\leq 0$$
$$-\frac{9}{2}\leq k\leq \frac{9}{2}$$
A: By C-S:
$$3=\sqrt{2x^2-4xy+6y^2}=\frac{2}{3}\sqrt{\left(2(x-y)^2+4y^2\right)\left(2+\frac{1}{4}\right)}\geq$$
$$\geq\frac{2}{3}\sqrt{(2(x-y)+y)^2}=\frac{2}{3}|2x-y|,$$ which gives
$$-\frac{9}{2}\leq2x-y\leq\frac{9}{2}.$$
The equality occurs for $$\left(\sqrt2(x-y),y\right)||\left(\sqrt2,\frac{1}{2}\right),$$ which with the condition gives
$$(x,y)=\left(\frac{5}{2},\frac{1}{2}\right)$$ for the right inequality and 
$$(x,y)=\left(-\frac{5}{2},-\frac{1}{2}\right)$$ for the left inequality,
which says that we got a maximal and the minimal value.
A: Let  $2x-y=z \implies y=2x-z$, then  $$f(x,y)=2x^2-4xy+6y^2-9=0 \implies z)+6(2x-z)=9 \implies 18x^2-20zx+6z^2-9=0 $$ Now for $x$ to be real $B^2\ge 4AC$
Then $$ 648-32 z^2 \ge 0 \implies  -9/2 \le z \le 9/2.$$
So the are the maximum and mininumm values of $2x-y$ are $\pm 9/2$
This also means that 2x-y=z will be tangent to the curve (ellipse) $f(x,y)=0$
when $z=\pm 9/2.$
