How do I go about solving this non-linear system of equations of 3 variables?

Question

In a problem on Lagrange multipliers, I need to find the maximum and minimum of, $$f(x, y) = \frac{1}{3}x^3 +\frac{1}{2}x^2y^2 + \frac{1}{3}y^3$$ subject to the constraint, $$g=x^2+y^2-2=0$$

Doing the method of Lagrange Multipliers as ${\nabla}f = {\lambda}{\nabla}g$ , (instead of finding the lagrangian $L$, and solving for $\frac{{\partial}L}{{\partial}x}, \frac{{\partial}L}{{\partial}y}, \frac{{\partial}L}{{\partial}\lambda}$),
I came to the problem of solving the 3 non-linear equations: $$x^2+xy^2=\lambda2x$$ $$y^2+x^2y = \lambda2y$$ $$x^2 +y^2-2=0$$Assuming I haven't gone wrong in forming these equations, how am I supposed to proceed from here to find all the constrained critical points? I have already found $(x,y) = (0,\pm\sqrt2)$ and $(x,y) = (\pm\sqrt2, 0)$, by factoring the first 2 equations etc. I am also aware that $(\pm1,\pm1)$ are solutions to the constraint equation, but there are more solutions that I haven't found.

How do I go about finding all solutions to these ugly non-linear equations?

Answer 1

You have already found all solutions where $x=0$ or $y=0$. So you can cancel a factor of $x$ from the first equation and $y$ from the second equation. You have $x + y^2 = 2\lambda$ and $y+ x^2 = 2\lambda$. So $x+y^2 = y + x^2$, or $x-y + y^2-x^2 = 0$. This factors as $(x-y)(1-x-y)$, so either $x=y$ or $x+y = 1$.

• If $x=y$, then the equation of constraint gives $2x^2=2$. So $(1,1)$ and $(-1,-1)$ are solutions.

• If $x+y = 1$, then the equation of constraint gives $x^2 + (1-x)^2 = 2$, or $2x^2 - 2x -1 = 0$. This gives the points $\left(\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2}\right)$ and $\left(\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}\right)$

Answer 2

$x^2+y^2-2=0$

Find extrema:$\ x^3/3+x^2y^2/2+y^3/3$

$\partial/\partial r\implies x^2\cos\theta+xy^2\cos\theta+yx^2\sin\theta+y^2\sin\theta=2r\lambda$

$\implies x^3+2x^2y^2+y^3=2(x^2+y^2)\lambda=4\lambda$

$\partial/ \partial \theta \implies x^2(-r\sin\theta)+xy^2(-r\sin\theta)+yx^2(r\cos\theta)+y^2r\cos\theta=0\lambda$

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

$=xy( x^2-x+1/4-1/4 -y^2+y -1/4+1/4)=xy( (x-1/2)^2-(y-1/2)^2)$

$=xy(x+y-1)(x-y)=0$

Alternatively:

Change of variables might help. Reduces it to a 1 D problem.

$x^2+y^2-2=0\implies r^2=2\implies r=\sqrt{2}$

$f(r=\sqrt{2},\theta)= (1/3)(2\sqrt{2})\cos^3\theta+(1/2)\sin^22\theta+(1/3)(2\sqrt{2})\sin^3\theta$

$df/d\theta=(2\sqrt{2})(-\sin\theta)\cos^2\theta+2\sin2\theta\cos2\theta+(2\sqrt{2})\cos\theta\sin^2\theta$

$0=\sin2\theta[-\sqrt{2}\cos\theta +2\cos^2\theta-2\sin^2\theta+\sqrt{2}\sin\theta]$

$0=2\sin2\theta[\cos^2\theta-\cos\theta/\sqrt{2}+(1/8)-(1/8)-\sin^2\theta+\sin\theta/\sqrt{2}]$

$0=2\sin2\theta[(\cos\theta-1/\sqrt{8})^2-(\sin\theta-1/\sqrt{8})^2]$

$0=2\sin2\theta(\cos\theta-\sin\theta)(\sin\theta+\cos\theta-1/\sqrt{2})$

$0=2\sin2\theta(\cos \theta - \sin \theta)(\sqrt{2}\sin(\theta+\pi/4)-\sqrt{2}/2)$

$2\theta=0\implies \theta \in \{0, \pi/2, \pi, 3\pi/2\}\implies (x,y) \in \{(\sqrt{2},0), (0,\sqrt{2}),(0,-\sqrt{2}), (-\sqrt{2},0)\}$

$(\cos\theta-\sin\theta)=0\implies \theta \in \{ \pi/4, 5\pi/4\}\implies (x,y)\in \{(1,1),(-1,-1)\}$

$(\sqrt{2}\sin(\theta+\pi/4)-\sqrt{2}/2)=0\implies \theta \in \{-\pi/12, 7\pi/12\}\implies (x,y)\in \{(\sqrt{1+\sqrt{3}/2},-\sqrt{1-\sqrt{3}/2}),(-\sqrt{1-\sqrt{3}/2},\sqrt{1+\sqrt{3}/2})\} $

Then test the 8 candidates.