I have doubts for what concerns the end of this exercise. I have the function $f(x, y, z) = x$ and I have to find min and max subject to the constraints $x^2+y^2+z^2 = \frac{5}{2}$ and $y + z = 1$.
I calculated the Jacobian matrix of the constraints and I found it has maximum rank of $2$, unless $x = y = z = 0$ thought this point doesn't nullify the constraints, hence they are regular. Thence the Lagrangian reads
$$L(x, y, z, \lambda, \mu) = x - \lambda(x^2+y^2+z^2-5/2) - \mu(y+z-1)$$
After a short algebra I found two solutions:
$$(\sqrt{2}, \frac{1}{2}, \frac{1}{2}) \qquad \text{for} \quad \lambda = \frac{1}{2\sqrt{2}}, \quad \mu = -\frac{1}{2\sqrt{2}}$$ $$(-\sqrt{2}, \frac{1}{2}, \frac{1}{2}) \qquad \text{for} \quad \lambda = -\frac{1}{2\sqrt{2}}, \quad \mu = \frac{1}{2\sqrt{2}}$$
*Question: I have been told that the sign of the multiplier determines if I found a max ($\lambda > 0$ or a min ($\lambda < 0$). I still did not really understand this well, but in any case how would I declare which of these is the max and which is the min?
The function over those points (where $y$ and $z$ are basically useless for $f$) is the same and it is $\sqrt{2}$.
Can you please help me in understanding how this work?
A couple of remarks, that hopefully are able to shed some light on your question(s)
I would strongly advise against using the sign of the Lagrange multiplier to check if its a minimum or maximum. Lets say you have the constraint $x=y+2$. Would you use $g=y-x+2$ or $g=x-y-2$?
You made a mistake with your second solution, it should read $(-\sqrt{2},\frac{1}{2},\frac{1}{2})$.
If you really want to make sure if you are dealing with a minimum, maximum or saddle you need to check the Hessian of the Lagrangian and its definiteness. https://en.wikipedia.org/wiki/Second_partial_derivative_test
In your case the first solution $(\sqrt{2},\frac{1}{2},\frac{1}{2})$ is the desired maximum and $(-\sqrt{2},\frac{1}{2},\frac{1}{2})$ is the desired minimum.