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I should find the minimum and maximum values of $f(x,y,z)=x+y+z$ given the constraints $x^2+y^2+z^2=1$ and $x−y-z=1$.

I found here a same exercise, but I don't know how the define the value of $x$, $y$, $z$.

Kuzja
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Evelin Danás
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    What is your question exactly...? Do you know how the method of Lagrange multipliers works? – StackTD May 17 '16 at 12:29
  • The question is what are the values of the x,y and z. Yes, I know.

    I tried to derivate the function accordance with x and y and z, but the values are 1 in every case, so I can't use lambda and mű. Secondy, I tried to express the value of x,y and z from the constraints, but because of both constraints contain x and y and z, so neither will loose. So I don't know how to do this exercise.

     – Evelin Danás May 17 '16 at 12:38

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This is a straightforward application of the LM method. Presumably you are happy to use that since you put it into the title. We put $f(x,y,z)=x+y+z-\lambda(x^2+y^2+z^2-1)-\mu(x-y-z-1)$ and set the partial derivatives wrt $x,y,z$ to 0. That gives: (1) $1-2\lambda x-\mu=0$, (2) $1-2\lambda y+\mu=0$, (3) $1-2\lambda z+\mu=0$.

(2) and (3) show that $y=z$. [Note that we cannot have $\lambda=0$ because then (1) and (3) give $1-\mu=1+\mu=0$, which is impossible.] So $x-y-z=1$ gives $x=2y$ and then substituting that in $x^2+y^2+z^2=1$ gives $y=0$ or $y=-\frac{2}{3}$, so we have the two solutions $(x,y,z)=(1,0,0),(-\frac{1}{3},-\frac{2}{3},-\frac{2}{3})$. We see that the first gives $x+y+z=1$ and the second gives $x+y+z=-\frac{5}{3}$. So the minimum value is $-\frac{5}{3}$ and the maximum value is $1$.

almagest
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