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Applying the Lagrange multiplier to a constrained optimization problem results in the system of equations below,

$$2xyz +xy+4yz+9zx=36$$ $$w=2yz+y+9z$$ $$w=2xz+x+4z$$ $$w=2xy+4y+9x$$

which I have trouble dealing with. I only need the values of $(x,y,z)$ in the end and could get them by on-line calculators. But, would really like to know the steps to solve the system. There are four sets of solutions, which indicates that one should get a quartic equation for any of the variables after elimination of other variables. But, it does not seem to be straightforward due to coupling among the variables, which makes the isolation procedure non-trivial.

Would like to get a helpful hand in seeing some light of it.

Quanto
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2 Answers2

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It seems that you want to find the conditional extrema of $x+y+z$ subject to the restriction in the first line. Fair enough.

You can solve the system using software, for instant with WolframAlpha, with the command GroebnerBasis, see here . This eliminates $x$, $y$, $z$ from the system, to get the equation that $w$ must satisfy: $w^6 - 252 w^4 - 1296 w^3=0$, that luckily factors as $w^3(w+6)(w+12)(w-18)=0$. Consider each case for $w$ in turn and solve the system in $x$, $y$, $z$. You can still do that using Groebner bases. For instance, in the case $w-18=0$, just add relation $w-18$, like here. It seems that all the solutions are nice...

orangeskid
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  • Your problem with non-negative variables we can solve by hand without LM. – Michael Rozenberg Feb 28 '20 at 06:04
  • @Michael Rozenberg: That would be interesting, why not post that as an answer – orangeskid Feb 28 '20 at 06:09
  • Because Quanto wants to solve this system. Also, my solution for non-negative variables it's something similar to the following: https://math.stackexchange.com/questions/3561383 By the way, there is a very ugly solution for the starting system. – Michael Rozenberg Feb 28 '20 at 06:15
  • @MichaelRozenberg - yeah, I'm more interested in solving the system than the root problem. Good solution in the link – Quanto Feb 28 '20 at 13:40
  • @orangeskid - Softwares, including the one I used, don't show the steps of elimination, which is what I could not manage. Do you have insight how the $w$-equation is actually derived. – Quanto Feb 28 '20 at 13:45
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Considering the system of equations $$2xyz +xy+4yz+9zx=36\tag 1$$ $$w=2yz+y+9z\tag 2$$ $$w=2xz+x+4z\tag 3$$ $$w=2xy+4y+9x\tag 4$$ what I did is the following

  • eliminate $y$ from $(2)$
  • eliminate $x$ from $(3)$
  • plug all of that in $(4)$ to end with $$\frac{(w+2) (2 w+9)}{(2 z+1)^2}-w-18=0 \implies z_\pm =\frac{\pm\sqrt{2 w^3+49 w^2+252 w+324}-w-18}{2 (w+18)}$$ Using $z_+$ and plugging in $(1)$ leads $$w^3-126 w-18 \left(\sqrt{(w+2) (w+18) (2 w+9)}+18\right)=0 $$ and the solutions are $w=-6$ and $w=18$. Using $z_-$ and plugging in $(1)$ leads $$-w^3+126 w+324-18 \sqrt{(w+2) (w+18) (2 w+9)}$$ and the solutions are $w=0$ and $w=-12$.

I hope no mistake.

Claude Leibovici
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  • Thanks. What do you mean by 'eliminate $y$ from (2)", which is not clear. What do you actually get from it? – Quanto Feb 28 '20 at 13:56
  • @Quanto. Solve for $y$ as a function of $(w,z)$ and do the same to get $x$. Then you have a quadratic in $z$ to get $z(w)$. Two roots. For each one, plug in $(1)$ and solve for $w$. Is this better ? Cheers :-) – Claude Leibovici Feb 28 '20 at 14:18
  • Okay, I'll take a shot along this path. – Quanto Feb 28 '20 at 15:27