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As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum. Lagrange multiplier gives a dirty calculation so I can't handle it. Is there any elegant way to find it? Thanks for any help.

p.s. Sorry. I make a typo in the $xy$-coefficient.

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    $x^2 + y^2 =0$ ? – 2017-01-21
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    @ramanujan It can't be 0. – 2017-01-21
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    Mathematica says that (barring me feeding it with the wrong system) the only candidate point is the obvious one with $x=y$ (involving cube roots of $53\pm6\sqrt{78}$). The system is, indeed, a bit hairy. That happens more often than not with Lagrange multipliers (much to the dismay of a teacher who wants to design paper&pencil solvable examples and exam questions). – 2017-01-21
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    Oh, sorry. I make a typo in the coefficient. I am going to fix it. – 2017-01-21
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    (after correcting the coefficient of the $xy$-term) If $x+y=1$ then $$1=(x+y)^3=x^3+3x^2y+3xy^2+y^3=(x^3+y^3)+3xy(x+y)=\cdots$$ In other words, that corrected constraint gives a straight line $x+y=1$. Surely you can manage... – 2017-01-21
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    The assumption that $x+y=1$ is not permissible because the equation is not homogeneous. – 2017-01-21
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    @DeepSea: You want it spelled out? $$(x^3+3xy+y^3-1)=(x+y-1)(x^2+y^2-xy+x+y+1).$$ That quadratic factor has no real zeros. – 2017-01-21
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    @JyrkiLahtonen $x^2+y^2-xy+x+y+1$ has solution of $x=y=-1$ – 2017-01-21
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    Thanks @IanMiller! I stand corrected. The "ellipse" shrinks to a single point. No wonder I didn't see it in the Mathematica ContourPlot. The quadratic form is obviously positive definite, so the zero locus of the quadratic is either an ellipse, the empty set, or a degenerate ellipse (as turned out to be the case here). – 2017-01-21

3 Answers 3

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Noting that $$x^3+y^3=(x+y)^3-3xy(x+y)$$ and $$(x+y)^3-1=(x+y-1)\left[(x+y)^2+(x+y)+1\right]$$ gives $$x^3+3xy+y^3=1 \\\\ \implies\ (x+y)^3-1-3xy(x+y)+3xy=0 \\\\ \implies\ (x+y-1)\left[(x+y)^2+(x+y)+1-3xy\right]=0 \\\\ \implies\ (x+y-1)(x^2+y^2-xy+x+y+1)=0$$ So either $x+y=1$ or $x^2+y^2-xy+x+y+1=0$.

Treating the latter as a quadratic equation in $x$ gives its discriminant as $(y-1)^2-4(y^2+y+1)=-3(y+1)^2$; so a real solution exists only when $y=-1$, at which value $x=-1$. This gives $x^2+y^2=(-1)^2+(-1)^2=2$.

As for $x+y=1$, we want the circle of $x^2+y^2=r$ $(r\ge0)$ to be tangent to that straight line. This occurs at $x=y=\dfrac12$, giving $r_{\min}=\left(\dfrac12\right)^2+\left(\dfrac12\right)^2=\dfrac12$.

Hence the minimum value is $$\min\left(2,\,\frac12\right)\ =\ \frac12$$

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$$x^3+3xy+y^3-1=(x+y-1)(x^2+y^2-xy+x+y+1)=0$$

So either $y=1-x$ or $(x,y)=(-1,-1)$

Subbing this in gives $x^2+(1-x)^2$ or $2$ to be minimized. Considering the first simple calculus gives us that $\frac{dy}{dx}=2x-2(1-x)=4x-2$. Setting this equal to zero gives $x=y=\frac{1}{2}$ as a possible minimum/maximum. Looking at the second derivative $\frac{d^2y}{dx^2}=4$ shows it is a minimum. Evaluating it gives $x^2+y^2=\frac{1}{2}$ which is less than the value at $(x,y)=(-1,-1)$ so the minimum value is $\frac{1}{2}$ at $(x,y)=\left(\frac{1}{2},\frac{1}{2}\right)$.

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Let $x=u+v$ and $y=u-v$. Then $x^2+y^2=2(u^2+v^2)$, while the equation $x^3+3xy+y^3=1$ becomes $2u^3+6uv^2+3(u^2-v^2)=1$, or

$$2u^3+3u^2-1=-3v^2(2u-1)$$

Factoring the cubic on the left hand side, we obtain

$$(u+1)^2(2u-1)=-3v^2(2u-1)$$

So either $2u-1=0$, in which case $v$ can be anything, or else $(u+1)^2=-3v^2$, which, because of the minus sign, is possible only if $u+1=v=0$. The latter possibility leads to

$$x^2+y^2=2((-1)^2+0^2)=2$$

while the former leads to

$$x^2+y^2=2\left(\left(1\over2\right)^2+v^2\right)={1\over2}+2v^2$$

which is clearly minimized when $v=0$. So the minimum value of $x^2+y^2$ is $1\over2$, and it occurs at $(x,y)=({1\over2}+0,{1\over2}-0)=({1\over2},{1\over2})$.