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Using Lagrange multipliers I obtained the following system of equations:

\begin{align*}x+y+z &= 20\\ x^2 + y^2 + z^2 &= 200\\ yz &= \lambda + 2 \mu x\\ xz &= \lambda + 2 \mu y\\ xy &= \lambda + 2 \mu z \end{align*}

I am struggling to solve this system of equations. I have managed to relate $\lambda$ and $\mu$. First I square the first equation and substitute the second equation:

$$(x+y+z)^2 = 200 + 2xy + 2xz + 2yz = 400 $$

$$xy+xz+yz = 100 $$

Then I add the last three equations

$$xy+xz+yz=3\lambda+2\mu(x+y+z) =3\lambda + 40 \mu=100$$

$$3 \lambda=100 -40\mu $$

But then I am stuck. I have confirmed that there are 6 distinct solutions to this system. What is the strategy when trying to solve this kind of system of equations?

Edit: The function I am trying to optimize is $xyz$ subject to the first two equations above.

s114
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3 Answers3

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The system is symmetric under permutations of $x,y$ and $z$, so a first step is to verify if there is a solution with $x=y=z$. One can easily verify that there isn't. Next, we try a solution which is partially symmetric, with $x=y\neq z$. The first two equations become \begin{align*} 2x+z&=20, \\ 2x^2+z^2&=200. \end{align*} Combining them we obtain a second degree equation for $x$,

$$3x^2-40x+100=0,$$

which has two real solutions, $x=10$ and $x=\frac{10}{3}$; the corresponding values of $z$ are $z=0$ and $z=\frac{40}{3}$, respectively. This yields a total of six solutions: $(10,10,0), (10,0,10), (0,10,10), (\frac{10}{3},\frac{10}{3},\frac{40}{3}), (\frac{10}{3},\frac{40}{3},\frac{10}{3}),$ and $(\frac{40}{3},\frac{10}{3},\frac{10}{3})$.

Finally, let's see if there are solutions with $x\neq y\neq z\neq x$. Subtracting the fourth from the third equation, we obtain $(y-x)z=2\mu(x-y)$. Since $x\neq y$, it implies $z=-2\mu$. If we now subtract the fifth from the third equation, we obtain $(z-x)y=2\mu(x-z)$. By the same reasoning, we conclude that $y=-2\mu$, but this contradicts the hypothesis that $y\neq z$. Therefore, there are no solutions with $x\neq y\neq z\neq x$.

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    +1 Nice. Also, you could notice that the function $f(t) = \lambda t + 2 \mu t^2$ has degree $\le 2$, and the solution satisfies $f(x)=f(y)=f(z)$ hence $x,y,z$ come from at most two values. So you could presume that the solution has the form $x,x,z$. – copper.hat Apr 17 '21 at 03:31
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The strategy is to eliminate variables, usually one at a time, "using up" one equation by solving it for an unknown, and then using the expression to remove all instances of the variable in the other equations. You slowly repeat the process until one remains, then you solve and back substitute.

Some approaches will definitely be more or less elegant than others!

Staring at your equations, I admit I don't see anything pretty. I'd probably start by eliminating $\lambda$ by setting it equal to $yz -2\mu x$ and substituting into the others, making

$xz = yz - 2 \mu x + 2 \mu y$ and $xy = yz-2\mu x +2 \mu z$. So now there are four equations in four unknowns. Next I would solve for $\mu$ in one of those and substitute into the other:

$xz - yz = 2 \mu (y-x)$ so $\mu = \frac{-z}{2}$, so plug that in, and you are down to three equations in three unknowns.

Continue in that fashion. The expressions will get messy, but you can get there systematically, removing one variable at a time. I'd probably go to the first equation next, and solve it for $x$ in terms of $y$ and $z$, et cetera...

The key is to be careful to remove all instances of one variable from each of the other equations, and not use that equation again until the back substitution stage at the end. At each step you should have one fewer equations and one fewer variables.

RobertTheTutor
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Suppose $x,y,z,\lambda,\mu$ satisfy the system of equations $$ \left\lbrace \begin{align*} x+y+z&=\,20&&(\text{eq}1)\\[4pt] x^2+y^2+z^2&=\,200&&(\text{eq}2)\\[4pt] yz&=\,\lambda+2\mu x&&(\text{eq}3)\\[4pt] xz&=\,\lambda+2\mu y&&(\text{eq}4)\\[4pt] xy&=\,\lambda+2\mu z&&(\text{eq}5)\\[4pt] \end{align*} \right. $$ First assume $x,y,z$ are all distinct.

Subtracting $(\text{eq}4)$ from $(\text{eq}3)$, we get $$ z(y-x)=-2\mu(y-x) $$ hence $z=-2\mu$.

But then, subtracting $(\text{eq}5)$ from $(\text{eq}4)$, we get $$ x(z-y)=-2\mu (z-y) $$ hence $x=-2\mu$, so $x=z$, contradiction.

Hence $x,y,z$ cannot be all distinct.

Next assume $x=y$.

Then from $(\text{eq}1)$ we get $z=20-2x$, so $(\text{eq}2)$ reduces to $$ 2(x-10)(3x-10)=0 $$ which yields the two qualifying triples \begin{align*} (x,y,z)&=\,(10,10,0)\\[4pt] (x,y,z)&=\, \left( \frac{10}{3}, \frac{10}{3}, \frac{40}{3} \right) \\[4pt] \end{align*} For each of the above triples, plugging in the values of $x,y,z$ into $(\text{eq}3),(\text{eq}4),(\text{eq}5)$, we get corresponding values for $\lambda,\mu$, yielding the two solutions \begin{align*} (x,y,z,\lambda,\mu)&=\,(10,10,0,100,-5)\\[4pt] (x,y,z,\lambda,\mu)&=\, \left( \frac{10}{3}, \frac{10}{3}, \frac{40}{3}, \frac{500}{9}, -\frac{5}{3} \right) \\[4pt] \end{align*} By symmetry, for the case $y=z$, we get the solutions \begin{align*} (x,y,z,\lambda,\mu)&=\,(0,10,10,100,-5)\\[4pt] (x,y,z,\lambda,\mu)&=\, \left( \frac{40}{3}, \frac{10}{3}, \frac{10}{3}, \frac{500}{9}, -\frac{5}{3} \right) \\[4pt] \end{align*} and similarly, for the case $z=x$, we get the solutions \begin{align*} (x,y,z,\lambda,\mu)&=\,(10,0,10,100,-5)\\[4pt] (x,y,z,\lambda,\mu)&=\, \left( \frac{10}{3}, \frac{40}{3}, \frac{10}{3}, \frac{500}{9}, -\frac{5}{3} \right) \\[4pt] \end{align*} so we have $6$ solutions in all.

quasi
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