Solving a system of equations in a Lagrange multiplier

Question

This question is with respect to the this answer of finding the maximum value of $z$ using Lagrange multipliers.

We get this set of equations;

$L(x,y,z,\lambda,\mu) = \begin{cases} x+y+z-8=0,\\2\,\mu\,x+\lambda=0,\\2\,\mu\,y+\lambda=0,\\2\,\mu\,z+\lambda+1=0,\\{x}^{2}+{y}^{2}+{z}^{2}-32=0. \end{cases}$

I solved the $2^{nd}$, $3^{rd}$ and $4^{th}$ equations getting;

$x= y =-\frac{\lambda}{2\mu}$, $z= -\frac{\lambda+1}{2\mu}$

I subbed the above into the $1^{st}$ and $5^{th}$ equations and got ;

$\frac{-2\lambda-\lambda-1}{2\mu}=8\implies -3\lambda-1=16\mu\implies16\mu+3\lambda=-1$

and,

$ \frac{\lambda^2}{4\mu^2}+\frac{\lambda^2}{4\mu^2} +\frac{\lambda^2+1+2\lambda}{4\mu^2}=32\implies 3\lambda^2+2\lambda +1= 128\mu^2\implies3\lambda^2+2\lambda-128\mu^2+1=0$

I thought the latter could be solved for $\lambda$ using the quadratic formlua;

$\lambda = \dfrac{-2\pm\sqrt{4-4(3)(1-128\mu^2)}}{6}\implies \lambda = \dfrac{-2\pm2\sqrt{768\mu^2-2}}{6}$

but it seems more complicated to me, so I left that train of thought.

I'm confused on how to proceed further , any hint , idea or suggestion would be highly appreciated.

Answer 1

I'd rather substitute $\mu$ into equation 5, noting $\mu=\frac{-1-3\lambda}{16}$, $$3\lambda^2+2\lambda-128(\frac{-1-3\lambda}{16})^2+1=0$$ $$3\lambda^2+2\lambda-0.5(1+3\lambda)^2+1=0$$ $$3\lambda^2+2\lambda-\frac{1}{2}(1+9\lambda^2+6\lambda)+1=0$$ $$-\frac{3}{2}\lambda^2-\lambda+\frac{1}{2}=0$$ We then have two $\lambda$'s, $\lambda=\frac{1}{3}$ or $\lambda=-1$.

You can check your answer on any "equation solver" on the web. Here's one.

Answer 2

Hint: we get $$x=-\frac{\lambda}{2\mu}=\frac{\frac{1}{3}+\frac{16}{3}\mu}{2\mu}$$ $$y=-\frac{\lambda}{2\mu}=\frac{\frac{1}{3}+\frac{16}{3}\mu}{2\mu}$$ $$z=\frac{-1-\lambda}{2\mu}=\frac{-1+\frac{1}{3}+\frac{16}{3}\mu}{2\mu}$$ plug these equation in $$x^2+y^2+z^2=32$$ and you will get $$\mu$$ Doing this we get $$-1/6\,{\frac { \left( -1+8\,\mu \right) \left( 8\,\mu+1 \right) }{{ \mu}^{2}}} =0$$