Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin. 
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
 A: I saw people recommending to use lagrange multiplier, but I know none of that. Instead, I tried this.
Firstly, you want to maximise/minimise (sqrt) $x^2+y^2$ with $x^4+y^4+3xy-2=0$. Consider $x^2+y^2=t$ and the intersection point $(u, v)$. i.e.
$$u^2+v^2=t, u^4+v^4+3uv-2=0$$
If at the point $(u, v)$, we can get both of the $\frac{dy}{dx}$ to be equal, then it means the curve $x^2+y^2-t=0$ and $x^4+y^4+3xy-2=0$ are tangent to each other.
Finding $\frac{dy}{dx}$ of both functions
Implicit differentiation gives
$$x^2+y^2=t \implies \frac{dy}{dx}=-\frac{x}{y}$$
$$x^4+y^4+3xy-2=0 \implies \frac{dy}{dx}=-\frac{4x^3+3y}{4y^3+3x}$$
We want to equal those. After simplifying
$$4xy^3+3x^2=4x^3y+3y^2$$
$$3(x^2-y^2)=4xy(x^2-y^2)$$
$$xy=\frac{3}{4} \textrm{ or } x=\pm y$$
If $xy=\frac{3}{4}$, $x^4+y^4+3(\frac{3}{4})-2=0 \implies x^4+y^4=-\frac{1}{4}$ which is impossible since LHS is non-negative.
Therefore, $x=\pm y$
**If ** $x=y$, $2x^4+3x^2-2=0\implies(2x^2-1)(x^2+2)=0\implies x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}, \sqrt{x^2+y^2}=1$
**If ** $x=-y$, $2x^4-3x^2-2=0\implies(2x^2+1)(x^2-2)=0\implies x=\pm \sqrt{2}, y=\mp \sqrt{2}, \sqrt{x^2+y^2}=2$
$$\therefore 1<\sqrt{x^2+y^2}<2$$
Side note: To the people doing lagrange multiplier, stop and calm down :) As mentioned in the comments, (apparently) this is the foundation of lagrange multiplier - Langrange multiplier is basically a generalisation of this to any constraint-equations and more variables :)
A: By Lagrange, we minimize
$$x^2+y^2+\lambda(x^4+y^4+3xy-2).$$
The derivatives on $x$ and $y$ yield
$$2x+\lambda(4x^3+3y)=2y+\lambda(4y^3+3x)=0,$$ and after elimination of $\lambda$,
$$2x(4y^3+3x)=2y(4x^3+3y)$$ or 
$$(8xy-6)(y^2-x^2)=0.$$
The solutions $y=\pm x$ imply
$$2x^4\pm3x^2-2=0,$$ giving
$$x^2+y^2=2x^2=\frac{\mp3\pm5}2$$ and only the positive solutions $1$ and $4$ are valid.
Next, with $xy=\dfrac34$,
$$x^4+y^4+3xy-2=(x^2+y^2)^2-2x^2y^2+3xy-2=(x^2+y^2)^2-\frac98+\frac94-2=0$$
and
$$x^2+y^2=\sqrt{\frac78}.$$ 
Anyway, we would have $(x-y)^2=x^2+y^2-2xy<0$, which is not possible.
Final answers: the solution points are $\pm\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$ and $\pm(\sqrt2,-\sqrt2).$
A: You've already got adequate answers, but this is just another way to do it -- especially fitted for this particular curve, as we'll find out.
The problem, reformulated, is to find the smallest circle intersecting the curve so that no point on the curve lies within the circle; on the other hand we need the largest circle intersecting the curve so that no point on the curve lies outside it.


*

*To find this least circle, let its radius be $r.$ Then we seek the least such $r$ satisfying the equations $x^2+y^2=r^2$ and $x^4+y^4+3xy=2.$ Squaring the first and substituting gives $$2-3xy+2(xy)^2=r^4,$$ a quadratic in $xy.$ Thus we easily find that the minimum value of $r^4,$ occurring at $xy=3/4,$ is given by $r^4=7/8.$ Hence our points lie on $xy=3/4$ and $x^2+y^2=\sqrt{7/8}.$ Thus we find that $x^2$ and $y^2$ are the roots of the quadratic $$m^2-\sqrt{7/8}m+9/16=0.$$ From here I believe you may proceed.

*To find the largest circle, we instead use the complementary circle of the hyperbola $x^2-y^2=s^2,$ whose vertices define the circle. Hence the radius of this circle is $s.$ Proceeding as above, we obtain that $$2-3xy-2(xy)^2=s^4,$$ which gives the maximum of $s^4$ as $25/8,$ occurring when $xy=-3/4.$ Thus our farthest points lie on the curves $xy=-3/4$ and $x^2-y^2=5/\sqrt 8.$ To solve this note that $(x^2+y^2)^2-(x^2-y^2)^2=4(xy)^2,$ which gives us that $x^2+y^2=\sqrt{43/8}.$ Hence $x^2$ and $y^2$ are the roots of the quadratic $$n^2-\sqrt{43/8}n+9/16=0,$$ from where you may proceed.
A: We'll prove that $\max(x^2+y^2)=4.$
Indeed, let $xy\geq0$.
Thus, by C-S $$2=x^4+y^4+3xy\geq x^4+y^4=\frac{(1+1)(x^4+y^4)}{2}\geq\frac{(x^2+y^2)^2}{2},$$
which gives $$x^2+y^2\leq2<4.$$
Now, let $xy\leq0.$
Thus, by C-S again and AM-GM we obtain:
$$2=x^4+y^4+3xy\geq\frac{(x^2+y^2)^2}{2}-\frac{3(x^2+y^2)}{2},$$
which gives $$(x^2+y^2)^2-3(x^2+y^2)-4\leq0$$ or
$$(x^2+y^2-4)(x^2+y^2+1)\leq0$$ or
$$x^2+y^2\leq4.$$
The equality occurs for $(x^2,y^2)||(1,1)$ and $x=-y,$ which with our condition gives:
$$(x,y)=(\sqrt2,-\sqrt2)$$ or
$$(x,y)=(-\sqrt2,\sqrt2).$$
By the similar way we can prove that $\min(x^2+y^2)=1,$ where the equality occurs for $$(x,y)=\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$$ or 
$$(x,y)=\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right).$$ 
