Example 3.40. Find the equation of the tangent planes to the surface $x^{2}+4 y^{2}+9 z^{2}=1$, parallel to the plane $x+y+2 z=1$.
Solution. Since the tangent plane must be parallel to the plane $x+y+2z=1$, the normal to the tangent plane must be collinear with the normal of the given plane. This means that, according to the condition of collinearity, the following condition holds:

$$
\frac{F_{x}^{\prime}(M)}{1}=\frac{F_{y}^{\prime}(M)}{1}=\frac{F_{z}^{\prime}(M)}{1},
$$

or $F_{x}^{\prime}=2x, F_{y}^{\prime}=8y, F_{z}^{\prime}=18z, \frac{2x}{1}=\frac{8y}{1}=\frac{18z}{2}$. From this, it follows that $2x=8y, 2x=9$ or $y=\frac{1}{4}x, z=\frac{2}{9}x$. The coordinates of the point $P\left(x, \frac{1}{4}x, \frac{2}{9}x\right)$ must satisfy the equation of the surface, since this point belongs to this surface. Substituting its coordinates into the equation of the surface, we get the following equation for $x: x^{2}+4 \cdot \frac{1}{16}x^{2}+9 \cdot \frac{4}{81}x^{2}=1$, from which $x= \pm \frac{6}{\sqrt{61}}$. Accordingly, $y= \pm \frac{3}{2 \sqrt{61}}, z= \pm \frac{4}{3 \sqrt{61}}$. Thus, we have two points of tangency $P_{1}\left(\frac{6}{\sqrt{61}}, \frac{3}{2 \sqrt{61}}, \frac{4}{3 \sqrt{61}}\right)$ and $P_{2}\left(-\frac{6}{\sqrt{61}},-\frac{3}{2 \sqrt{61}},-\frac{4}{3 \sqrt{61}}\right)$ and, accordingly, two tangent planes, which, according to (3.30), are described by the equations:

$$
\begin{gathered}
1 \cdot\left(x \pm \frac{6}{\sqrt{61}}\right)+1 \cdot\left(y \pm \frac{3}{2 \sqrt{61}}\right)+2 \cdot\left(z \pm \frac{4}{3 \sqrt{61}}\right)=0 \Rightarrow \\
\Rightarrow x+y+2z \pm \frac{109}{6 \sqrt{61}}=0
\end{gathered}
$$

## 3.7. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

Let the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ be defined in some domain $G \subset E^{n}$ and let the point $P_{0} \in G$. The value of the function at this point $f\left(P_{0}\right)$ is called a local maximum (local minimum) of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ in the domain $G$ if there exists some neighborhood $U\left(P_{0}\right) \subset G$ of the point $P_{0}$ such that for all points $P \in U\left(P_{0}\right) P_{0}$ the inequality $f(P)f\left(P_{0}\right)\right)$ holds. The maximum or minimum of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is also called a local extremum of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ in the domain $G$.

Remarks. 1) A neighborhood of the point $P_{0}$ is understood to be any region containing the point $P_{0}$. 2) The notation $U\left(P_{0}\right) P_{0}$ means a punctured neighborhood of the point $P_{0}$, that is, a neighborhood of this point without the point itself. 3) From the definition of a local extremum of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ at the point $P_{0}\left(x_{1}^{0}, x_{2}^{0}, \ldots, x_{n}^{0}\right)$, it follows that $f\left(P_{0}\right)$ is the largest or smallest value of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ in the neighborhood of the point $P_{0}$, but it is not, in general, the largest or smallest value of the function in the domain $G$, that is, a global extremum.

Necessary Condition for an Extremum

If the point $P_{0}$ is a point of local extremum of $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ and if the first-order partial derivatives of the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ exist at the point $P_{0}$, then all these partial derivatives are zero, that is,

$$
f_{x_{i}}^{\prime}\left(P_{0}\right)=0,(i=1,2, \ldots, n)
$$

Points where the first-order partial derivatives are zero are called critical, or stationary, or points of possible extremum.

Remarks. 1) The function $f(P)$ may have an extremum at points where the partial derivatives do not exist. 2) Condition (3.34) is not a sufficient condition for a local extremum. For example, the partial derivatives of the function $z=y^{2}-x^{2}$
are zero at the point $O(0,0)$, but this function does not have an extremum at the specified point, since it equals zero there and the inequality $y^{2}-x^{2}0\right)$ does not hold in any neighborhood of the point $(0,0)$.