Local extrema of the implicit function defined by $5x^2+5y^2+5z^2-2xy-2xz-2yz-72=0$ A problem asked us to determine the local extrema of an implicit function $z=z(x,y)$ defined by the equation $5x^2+5y^2+5z^2-2xy-2xz-2yz-72=0$. My instructor went about this as follows:
Let $F:\mathbb{R}^3 \to \mathbb{R}, F(x, y, z)=5x^2+5y^2+5z^2-2xy-2xz-2yz-72$. Then, $F$ defines explicitely the function $z=z(x,y)$ in the neighborhood of a point $(x_0, y_0, z_0) \in \mathbb{R}^3$ if $F(x_0, y_0, z_0)=0$ and $\frac{\partial F}{\partial z}(x_0, y_0, z_0) \ne 0$ (we wish to use the implicit function theorem, I forgot to mention that). After writing the conclusion of the implicit function theorem, he drew the conclusion that we need to solve the system
$\begin{cases}
F(x,y,z)=0 \\
\frac{\partial F}{\partial z}(x, y, z)\ne 0 \\ 
\frac{\partial F}{\partial x}(x, y, z)=0 \\
\frac{\partial F}{\partial y}(x, y, z)=0 
\end{cases}
$ in order to find the critical points of our implicit function.
I don't really understand how we get this system. I understand that the first two equations come from the fact that we are searching for those points $(x_0, y_0, z_0)$ where we can apply the implicit function theorem. My instructor said that the last two equations are there  because we want to have $\frac{\partial z}{\partial x}(x, y)=0$ and $\frac{\partial z}{\partial y}(x, y)=0$, but I don't understand why this is the case (I know the formula for implicit differentiation, but I still don't get it). Furthermore, why should all these $4$ equations hold simultaneously? I don't understand why the point for which we apply the implicit function theorem should necessarily also be a critical point for our function $z(x, y)$.
 A: You can reach the result without using a multivariable calculus. Observe that
$$\begin{align}&f(x,y,z)=5x^2+5y^2+5z^2-2xy-2xz-2yz-72 &\\
&=\frac 15(5 x - y - z)^2 + \frac{3}{10}(4 y - z)^2 + \frac {9}{2} (z^2 - 16)\end{align}$$
So, for global and local minimum we need to take
$$z=0, ~ 4y-z=0,~ 5x-y-z=0$$
$$\implies x=y=z=0$$
Thus, we get
$$\text{min}\left\{f(x,y,z)\right\}=-72 \\ ~\text{at} ~ x=0, y=0, z=0$$
The polynomial $$\begin{align} f(x,y,z)=\frac 15(5 x - y - z)^2 + \frac {3}{10} (4 y - z)^2 + \frac {9}{2} (z^2 - 16)\end{align}$$
shows that, local and global maxima doesn't exist.
A: Without advanced calculus.
As $f(x,y,z) = 5 x^2 + 5 y^2 + 5 z^2 - 2 x y - 2 x z - 2 y z - 27 = 0$ represents a surface in $\mathbb{R}^3$ more precisely an ellipsoid, if we want to know taking as reference coordinate the $z$ axis, what the extremals regarding the $z$ axis, then this quest can be explained as: find the points in which that surface is tangent to the generic plane $z = \lambda$ or
$$
f(x^*,y^*,\lambda) = 0\Rightarrow x^* = \frac{1}{5} \left(y^*+\lambda\pm\sqrt{3} \sqrt{4 y^* \lambda-8 (y^*)^2-8 \lambda^2+45}\right)
$$
but at tangency this point should be unique so
$$
4 y^* \lambda-8 (y^*)^2-8 \lambda^2+45=0
$$
solving now for $y^*$ we have
$$
y^* = \frac{1}{4} \left(\sqrt{15} \sqrt{6-\lambda^2}+\lambda\right)
$$
$y^*$ should also be unique then $\lambda = z^* = \pm\sqrt{6}$ so the extremals are located at
$$
\left[
\begin{array}{ccc}
x^* & y^* & z^* \\
 -\frac 12\sqrt{\frac{3}{2}} & -\frac 12\sqrt{\frac{3}{2}} & -\sqrt{6} \\
 \frac 12\sqrt{\frac{3}{2}} & \frac 12\sqrt{\frac{3}{2}} & \sqrt{6} \\
\end{array}
\right]
$$
Now with calculus
Taking $f(x,y,z(x,y))= 5 x^2 + 5 y^2 + 5 z^2(x,y) - 2 x y - 2 x z(x,y) - 2 y z(x,y) - 27 = 0$ we have
$$
f_x = 10 x+10z z_x-2y-2z-x z_z -2y z_x = 0
$$
and then
$$
z_x = \frac{5x-y-z}{x+y-5z}
$$
analogously
$$
f_y = 10y+10 z z_y -2x-2xz_y -2z-2yz_y = 0
$$
with
$$
z_y = \frac{5y-x-z}{x+y-5z}
$$
then the stationary  points for  $z(x,y)$ verify
$$
\cases{z_x = 0\Rightarrow 5x-y-z = 0\\
z_y = 0 \Rightarrow 5y-x-z=0}
$$
Now the intersection of the ellipsoid with the line defined as the intersection of two planes, gives the solution
$$
\cases{
5 x^2 + 5 y^2 + 5 z^2 - 2 x y - 2 x z - 2 y z - 27 = 0\\
5x-y-z = 0\\
5y-x-z=0
}
$$
A: We need to find the local extrema of the function $z(x,y)$ defined by $F(x,y,z(x,y))=0$. And here belows is the explication why you need to have
\begin{cases}
F(x,y,z) = 0 \\
\frac{\partial F(x,y,z)}{\partial x} = 0 \\
\frac{\partial F(x,y,z)}{\partial y} = 0\\
\frac{\partial F(x,y,z)}{\partial z} \ne 0  \tag{*}
\end{cases}
I suppose you know the method for local extrema of explicit function $z(x,y)$. The idea is to calculate and solve
$$\frac{\partial z(x,y)}{\partial x} = 0 \tag{1}$$
$$\frac{\partial z(x,y)}{\partial y} = 0$$
From the the chain rule derivative
$$\frac{\partial F(x,y,z(x,y))}{\partial x} =  \frac{\partial F(x,y,z)}{\partial x}+\frac{\partial F(x,y,z)}{\partial z}\frac{\partial z(x,y)}{\partial x} \tag{2}$$
As $ F(x,y,z(x,y)) = 0$ then the left hand side of $(2)$ is equal to $0$.
$$(2) \implies \frac{\partial F(x,y,z)}{\partial x}+\frac{\partial F(x,y,z)}{\partial z}\frac{\partial z(x,y)}{\partial x} =0 \tag{3}$$
We notice that $\frac{\partial F(x,y,z)}{\partial z} \ne 0$ (the forth equation of $(\color{blue}*)$) because if not, $z(x,y)$ can't be defined.
So, from $(3)$, we have
$$\frac{\partial z(x,y)}{\partial x} = - \frac{\frac{\partial F(x,y,z)}{\partial x}}{\frac{\partial F(x,y,z)}{\partial z}}$$
and hence, from $(1)$, we have
$$- \frac{\frac{\partial F(x,y,z)}{\partial x}}{\frac{\partial F(x,y,z)}{\partial z}} = 0$$
or the second equation of $(\color{blue}*)$
$$\frac{\partial F(x,y,z)}{\partial x} = 0$$
By the same method, we need also the third equation of $(\color{blue}*)$
$$\frac{\partial F(x,y,z)}{\partial y} = 0$$
Conclusion: In order to find the extrama of implicit function $F(x,y,z) = 0$, we need to solve $(\color{blue}*)$
\begin{cases}
F(x,y,z) = 0 \\
\frac{\partial F(x,y,z)}{\partial x} = 0 \\
\frac{\partial F(x,y,z)}{\partial y} = 0\\
\frac{\partial F(x,y,z)}{\partial z} \ne 0
\end{cases}
Remark:
$$
(\color{blue}*) \iff
\begin{cases}
\frac{\partial z(x,y)}{\partial x} = 0  \\
\frac{\partial z(x,y)}{\partial y} = 0
\end{cases}
$$
A: Since
$$f(x,y,z)=(x+y+z)^2+2(x-y)^2+2(y-z)^2+2(z-x)^2-72,$$
then the global minimum achieves if
$$(x=y=z)\wedge(x+y+z=0),$$
i.e.
$$\min\limits_{\large x,y,z\,\in \mathbb R} f(x,y,z)=f(0,0,0)=-72.$$
There is a single extremum, because the second-order curve is a unimodal one.
