$s \in \mathbb{R}$ and $s > 0$.
$S = \{(x, y, z) | 0 < x < s, 0 < y < s, 0 < z < s, x + y + z = 2s\}$.
$f(x, y, z) = (s - x) (s - y) (s - z)$
Find all the points $(x, y, z) \in S$ that maximaize the value of $f$ by using Lagrange Multiplier Method.
Let $T =\{(x, y, z) | 0 \leq x \leq s, 0 \leq y \leq s, 0 \leq z \leq s, x + y + z = 2s\}$.
$T$ is a compact set and $f$ is continuous.
So, $f$ has a maximum on $T$.
$(\frac{2}{3} s, \frac{2}{3} s, \frac{2}{3} s) \in T$ and $f(\frac{2}{3} s, \frac{2}{3} s, \frac{2}{3} s) = \frac{s^3}{27} > 0$.
So, the maximum value is larger than $0$.
If $x = s$ or $y = s$ or $z = s$, then $f = 0$.
If $x = 0$ or $y = 0$ or $z = 0$, then $f = 0$.
So, if $(x_0, y_0, z_0)$ is a maximum point on $T$, then $(x_0, y_0, z_0) \in S$.
So, $f$ has a maximum on $S$.
Let $g(x, y, z) = x + y + z - 2s$.
$g = 0$.
$f_x = \lambda g_x$.
$f_y = \lambda g_y$.
$f_z = \lambda g_z$.
Solving this equation, we have $(x, y, z) = (\frac{2}{3} s, \frac{2}{3} s, \frac{2}{3} s)$.
$(\frac{2}{3} s, \frac{2}{3} s, \frac{2}{3} s)$ is the only candidate for local maximum or local minimum.
What should I do next?
If $(x_0, y_0, z_0)$ is a maximum point on $S$,
then
$f_x(x_0, y_0, z_0) = \lambda g_x(x_0, y_0, z_0)$
$f_y(x_0, y_0, z_0) = \lambda g_y(x_0, y_0, z_0)$
$f_z(x_0, y_0, z_0) = \lambda g_z(x_0, y_0, z_0)$
for some $\lambda$.
Is this true?