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I'm solving a problem where I have to maximize a function according to 2 restrictions, so in the end I got to the following system of 5 equations:

$$ \begin{cases} 2x &= 2x \lambda - 4\lambda +\mu \\ 2y &= 2y \lambda - 4\lambda +\mu \\ 2z &= - \mu \\ 0 &= x^2 -4x +y^2-4y \\ 0 &= x+y-z \end{cases} $$

I substracted the first two to get:

$$ \begin{align} 2x-2y=2x\lambda-2y\lambda \\ 2(x-y)=2\lambda(x-y)\\ 2(x-y)-2\lambda(x-y)=0 \\ 2(x-y)(1-\lambda)=0 \end{align} $$ So from this I get that either $\lambda=1$ or $x=y$.

I "explored" the $x=y$ path by using the fourth and fifth equations: $$ \begin{align} x^2-4x+x^2-4x=0 \\ 2x^2-8x=0 \end{align} $$ From this I get that either $x=4$ or $x=0$ and, since $x=y$ then also $y=4$ or $y=0$. I used the fifth equation to get the value of $z$: $$ \begin{align} x+y-z=0 \\ z=x+y \\ \end{align} $$ So now $z=0+0=0$ or $z=4+4=8$.

So far so good. But I'm not sure about how to explore the $\lambda=1$ path, as the only equations that have lambda in it have already been used, well, to get that $\lambda=1$. Can I "reuse" them?

Also, I completely eliminated $\mu$ and never found any possible values for it. Is that correct?

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You need to consider $\lambda =1$ in the first two equations to get your $\mu$ and use it to find z accordingly.

In that case you have used $\mu$ which helps you to find new points.

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  • $\begingroup$ So this means I can reuse the first two equations, which I already used to find the value of $\lambda$? Is that a valid "move"? $\endgroup$ – Floella Jun 7 '18 at 23:02
  • $\begingroup$ Ok, so if I do this: $2x=2x-4+\mu$ then I get that $\mu=4$ and from the third equation follows that $z=-2$. But now I'm stuck with getting values for $x$ and $y$... $\endgroup$ – Floella Jun 7 '18 at 23:17
  • $\begingroup$ With z=-2 you can solve the last two equations for x and y. $\endgroup$ – Mohammad Riazi-Kermani Jun 8 '18 at 0:49

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