I need do find global max and min of $$ z=x+y^2 $$ inside (or on the border of) the volume described by: $$ x^2+y^2-25=0 $$
I've already found maximum values in $(\frac{1}{2}, +\frac{3\sqrt{11}}{2}, \frac{101}{4})$ and $(\frac{1}{2}, -\frac{3\sqrt{11}}{2}, \frac{101}{4})$ by writing the system of equations of the above ones, getting the derivative with respect to $z$ and finally finding the zeroes.
But I cannot find the minimum using some algebraic method$(-5, 0, -5)$.
I've seen from this answer that I should check the extreme values of $x$, but I do not understand the reason why, shouldn't these values be already included thanks to the system of equations?
(Do not use Lagrange multipliers nor Hessian matrix)
In the univariate case where you want to find the global extrema of some differentiable function $f(x)$ on some interval $x \in [a,b]$, would you not also calculate $f(a)$ and $f(b)$ to determine whether the extrema occur at the endpoints?
For instance, if I asked you for the global extrema (min and max) of $$f(x) = 2x^2 - 4x + 1$$ on the interval $x \in [-1,5]$, you would calculate $$f'(x) = 4x - 4,$$ hence the point $x = 1$ is a critical point, and $f(1) = 2(1)^2 - 4(1) + 1 = -1$. The second derivative is $$f''(1) = 4 > 0,$$ so $x = 1$ is a local minimum. But now we must also calculate $$f(-1) = 2(-1)^2 - 4(-1) + 1 = 7, \\ f(5) = 2(5^2) - 4(5) + 1 = 31.$$ And we see that $f(5)$ is the global maximum on this interval, even though neither $-1$ nor $5$ is a critical point. For the same exact function $f$, it is also possible for the global extrema to be different than what we found above, if the interval on which the extrema are to be found is different than $[-1,5]$.
The same idea applies to your situation. On the boundary, which is the circle $x^2 + y^2 = 25$, the function may have some global extremum for which the partial derivatives are not zero--i.e., it is not a critical point. Conversely, there may also be a global extremum strictly inside the circle, but in this case, we know that if the function is smooth, then such an extremum will be a critical point.
Now that we have established this conceptual framework, consider how we can proceed. We would search for critical points by noting that $$\frac{\partial z}{\partial x} = 1, \quad \frac{\partial z}{\partial y} = 2y,$$ hence there are no critical points inside the circle since the partial with respect to $x$ can never be zero. Any global extremum must therefore be a boundary point. In this case, the boundary is easily parametrized, which lets us reduce the question to differentiation of a single variable:
$$(x,y) = (5 \cos \theta, 5 \sin \theta)$$
is the polar representation of a point on the boundary $x^2 + y^2 = 25$, hence $$z = 5 \cos \theta + 25 \sin^2 \theta$$ on this boundary, and we can find the extrema in terms of the angle $\theta \in [0,2\pi)$. Differentiating, $$\frac{dz}{d\theta} = -5 \sin \theta + 50 \cos\theta \sin \theta = 5 \sin \theta (-1 + 10 \cos \theta).$$ So the critical points must satisfy $$\sin \theta = 0 \quad \text{or} \quad \cos \theta = \frac{1}{10}.$$ In the first case, this implies $\theta \in \{0, \pi\}$, and in the second, $$\theta = \left\{\arccos \frac{1}{10}, 2\pi - \arccos \frac{1}{10} \right\}.$$ In terms of $(x,y)$ coordinates, these are $$(x,y) \in \left\{(5,0), (-5,0), \left(\frac{1}{2}, \frac{3\sqrt{11}}{2}\right), \left(\frac{1}{2}, -\frac{3\sqrt{11}}{2}\right)\right\}.$$ The $z$-values are therefore $$(x,y,z) \in \left\{(5,0,5), (-5,0,-5), \left(\frac{1}{2}, \frac{3\sqrt{11}}{2}, \frac{101}{4}\right), \left(\frac{1}{2}, -\frac{3\sqrt{11}}{2}, \frac{101}{4}\right)\right\}.$$ Since the $z$-coordinate of the latter two are greater than the first, we have shown that the global minimum is the second point $(-5,0,-5)$ and the global maxima are at the third and fourth points.
Alternatively, a more simple solution can be found in this specific case by observing that if $x^2 + y^2 \le 25$, then $$0 \le y^2 \le 25 - x^2,$$ hence $$x \le z = x + y^2 \le 25 + x - x^2.$$ So the extrema of $z = x + y^2$ on the closed disk must also satisfy this inequality. Since $x \in [-5,5]$, we see that when $x = -5$, $$-5 \le z \le 25 + (-5) - (-5)^2 = -5,$$ meaning that $z = -5$ is the global minimum, since any other choice of $x$ will lead to $z$ being greater than $-5$. As for the global maximum, we want to make $25 + x - x^2$ as large as possible, and this can be done by completing the square: $$25 + x - x^2 = -(x^2 - x - 25) = -\left( (x - \tfrac{1}{2})^2 - \tfrac{1}{4} - 25 \right) = \tfrac{101}{4} - (x - \tfrac{1}{2})^2.$$ Since no real square is negative, the largest possible value is attained when $x - \frac{1}{2} = 0$, or $x = \frac{1}{2}$, which corresponds to $z \le \frac{101}{4}$. All that is left is to check that this value is attainable for some choice of $y$: when $x = \frac{1}{2}$, we have $$y^2 \le 25 - x^2 = \frac{99}{4},$$ hence $y \in \pm \frac{3\sqrt{11}}{4}$, and $z = x+y^2 = \frac{1}{2} + \frac{99}{4} = \frac{101}{4}$. So we have found the global maxima this way as well. Note this method has completely avoided the use of calculus.