2 constraint optimization (Lagrange multipliers)

Question

Determine the critical points of $x^3 + y^3 +z^3$, such that $x^2 + y^2 +z^2 = 1$ and $x + y+ z = 0$ by hand.

Attempt at a solution: I seem to figure out $-1$ as the multiplier for $x+y+z=0$ but can't proceed as the equations seems duplicated.

Answer 1

Set $f(x,y,z)=x^3+y^3+z^3$, $u(x,y,z)=x^2+y^2+z^2-1$ and $v(x,y,z)=x+y+z$. Then the critical points of $f$ restrict to $u(x,y,z)=0$ and $v(x,y,z)=0$ is too the critical points of $$ F(x,y,z,\mu,\lambda)=f(x,y,z)+\mu u(x,y,z)+ \lambda v(x,y,z). $$ The critical points of $F$ satisfies $\nabla F(x,y,z,\mu,\lambda )=0$.That is, $$ \begin{array}{lcrcr} \left\{ \begin{array}{ccc} (1)&\dfrac{\;\partial}{\partial {x}}F(x,y,z)&=&0\\ (2)&\dfrac{\;\partial}{\partial {y}}F(x,y,z)&=&0\\ (3)&\dfrac{\;\partial}{\partial {z}}F(x,y,z)&=&0\\ (4)&\dfrac{\;\partial}{\partial {\mu}}F(x,y,z)&=&0\\ (5)&\dfrac{\;\partial}{\partial {\lambda}}F(x,y,z)&=&0\\ \end{array} \right. & \Longleftrightarrow & \left\{ \begin{array}{crcl} (1)& 3x^2+2\mu x+\lambda&=&0\\ (2)& 3y^2+2\mu y+\lambda&=&0\\ (3)& 3z^2+2\mu z+\lambda&=&0\\ (4)& x^2+y^2+z^2&=&1\\ (5)& x+y+z&=&0\\ \end{array} \right. \end{array} $$ Note that $$ (1)+(2)+(3) : 3(x^2+y^2+z^2)+2\mu(x+y+z)+3\lambda=0\implies \lambda=-1.\\ (1), \mbox{ and } \lambda=-1 \implies x=\frac{-2\mu\pm \sqrt{4\mu^2+12}}{6}\\ (2), \mbox{ and } \lambda=-1 \implies y=\frac{-2\mu\pm \sqrt{4\mu^2+12}}{6}\\ (3), \mbox{ and } \lambda=-1 \implies z=\frac{-2\mu\pm \sqrt{4\mu^2+12}}{6}\\ $$ Cases $$ \begin{array}{|c|c|c|c|c|} \hline & x & y & z & \mu \mbox{ in (5) } \\\hline Case 1 & x=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & \mbox{ not existis } \\\hline Case 2 & x=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 3 & x=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 4 & x=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 5 & x=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 6 & x=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 7 & x=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu+ \sqrt{4\mu^2+12}}{6} & \mbox{ existis } \\\hline Case 8 & x=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & y=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & z=\frac{-2\mu- \sqrt{4\mu^2+12}}{6} & \mbox{ not existis } \\\hline \end{array} $$ For cases $2$ to $7$ we have in $(5) x+y+z=0$ \begin{align} -\mu\pm \frac{\sqrt{4\mu^2+12}}{6}=0 \implies & -\mu =\mp \frac{\sqrt{4\mu^2+12}}{6} \\ \implies & (-\mu )^2=\left(\mp \frac{\sqrt{4\mu^2+12}}{6}\right)^2 \\ \implies & 36\mu^2=|4\mu^2+12| \\ \implies & 36\mu^2=4\mu^2+12 \\ \implies & 32\mu^2=12 \\ \implies &\mu^2=\frac{3}{8} \\ \implies &\mu=\sqrt{\frac{3}{8}}=\frac{\sqrt{3}}{2\sqrt{2}} \\ \implies &\mu=\frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{4} \\ \end{align} The expression $\sqrt{4\mu^2+12}$ for $\mu=3/8$ is $$ \sqrt{4\mu^2+12}=\sqrt{27/2} =3\sqrt{3/2}=3\frac{\sqrt{6}}{2} $$ Then the expressions to $x$, $y$ and $z$, in case 2, for exemple, are $$ \begin{array}{c} x=\frac{1}{6}\left(-2\frac{\sqrt{6}}{4}+ 3\frac{\sqrt{6}}{2} \right) =\frac{\sqrt{6}}{6} \\ y=\frac{1}{6}\left(-2\frac{\sqrt{6}}{4}+ 3\frac{\sqrt{6}}{2} \right) =\frac{\sqrt{6}}{6} \\ z=\frac{1}{6}\left(-2\frac{\sqrt{6}}{4}- 3\frac{\sqrt{6}}{2} \right) =-2\frac{\sqrt{6}}{6} \end{array} $$

Answer 2

HINT: use $$F(x,y,z,\alpha,\beta)=x^3+y^3+z^3+\alpha(x^2+y^2+z^2-1)+\beta(x+y+z)$$ and differentiate $F$ with respect to $x,y,z,\alpha,\beta$ we get $$\frac{\partial F}{\partial x}=3x^2+2\alpha x+\beta=0$$ $$\frac{\partial F}{\partial y}=3y^2+2\alpha y+\beta=0$$ $$\frac{\partial F}{\partial z}=3z^2+2\alpha z+\beta=0$$ $$\frac{\partial F}{\partial \alpha}=x^2+y^2+z^2=1$$ $$\frac{\partial F}{\partial \beta}=x+y+z=0$$ Have you solved this system?