Example 3. At what point is the tangent plane to the elliptic paraboloid $z=2 x^{2}+4 y^{2} \quad$ parallel to the plane $8 x-32 y-2 z+3=0$? Write the equations of the normal and the tangent plane at this point.
Solution. In this case, the surface is given by an equation solved for $z$, i.e., an equation of the form (8.5), where $f(x, y)=2 x^{2}+4 y^{2}$.

The equation of the tangent plane at any point $M$ according to formula (8.6) is

$$
z-z_{0}=4 x_{0}\left(x-x_{0}\right)+8 y_{0}\left(y-y_{0}\right)
$$

or

$$
4 x_{0}\left(x-x_{0}\right)+8 y_{0}\left(y-y_{0}\right)-\left(z-z_{0}\right)=0.
$$

This plane, by the condition of the problem, must be parallel to the plane $8 x-32 y-2 z+3=0$. Since the planes are parallel, their coefficients of the corresponding current coordinates are proportional:

$$
\frac{4 x_{0}}{8}=\frac{8 y_{0}}{-32}=\frac{-1}{-2}
$$

from which

$$
\frac{1}{2} x_{0}=\frac{1}{2}, -\frac{1}{4} y_{0}=\frac{1}{2}
$$

i.e.,

$$
x_{0}=1, y_{0}=-2
$$

From the equation $z=2 x^{2}+4 y^{2}$, we find $z_{0}=18$. Therefore, $M_{0}(1,-2,18)$ is the desired point.

Equation (A) with the obtained values of $x_{0}, y_{0}, z_{0}$ takes the form

$$
4(x-1)-16(y+2)-(z-18)=0
$$

or

$$
4 x-16 y-z-18=0
$$

According to formula (8.7), we write the equations of the normal at the point $M_{0}(1,-2,18)$:

$$
\frac{x-1}{4}=\frac{y+2}{-16}=\frac{z-18}{-1}
$$

## Problems

Write the equations of the normal and the tangent plane to each of the following surfaces at the specified point:

1. $2 x^{2}+3 y^{2}+4 z^{2}=9, M_{0}(1,-1,1)$.
2. $x^{2}+y^{2}-z^{2}=0, M_{0}(3,4,5)$.
3. $z=4 x^{2}-9 y^{2}, M_{0}(1,1,-5)$.
4. $x=2 y^{2}+3 z^{2}, M_{0}(5,1,1)$.
5. At which points of the surface

$$
x^{2}+y^{2}+z^{2}+8 x-4 y+2 z+12=0
$$

are the tangent planes parallel to the coordinate planes?

6. To the surface $x^{2}+y^{2}+4 z^{2}=4$, draw a tangent plane that cuts off equal segments on the coordinate axes.

Answers

1. $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-1}{4} ; 2 x-3 y+4 z-9=0$. 2. $\frac{x-3}{3}=\frac{y-4}{4}=\frac{z-5}{-5}$; $3 x+4 y-5 z=0$. 3. $\frac{x-1}{8}=\frac{y-1}{-18}=\frac{z+5}{-1} ; 8 x-18 y-z+5=0$. 4. $\frac{x-5}{-1}=$
$=\frac{y-1}{4}=\frac{z-1}{6} ; x-4 y-6 z+5=0$. 5. At points $M_{1}(-7,2,-1)$, $M_{2}(-1,2,-1)$ the tangent planes are parallel to $O y z$; at points $M_{3}(-4,5,-1)$, $M_{4}(-4,-1,-1)$ the planes are parallel to $O x z$; at points $M_{5}(-4,2,2)$, $M_{6}(-4,2,-4)$ the planes are parallel to $O x y$. 6. $x+y+z-3=0 ; x+y+z+3=0$.

## § 8.2. Extremum of a Function of Several Variables

The maximum (minimum) of a function $z=f(x, y)$ is a value $f\left(x_{0}, y_{0}\right)$ that is greater (less) than all other values it takes in points sufficiently close to the point $M_{0}\left(x_{0}, y_{0}\right)$ and different from it.

The maximum or minimum of a function is called its extremum. The point at which the extremum is achieved is called the extremum point.

Similarly, the extremum of a function of three or more variables is defined.

The extremum of a function of several variables can only be achieved at points lying inside the domain of its definition, where all first-order partial derivatives are zero. Such points are called critical points. For a function of two variables $z=f(x, y)$, critical points are found from the system of equations:

$$
\left.\begin{array}{l}
f_{x}^{\prime}(x, y)=0 \\
f_{y}^{\prime}(x, y)=0
\end{array}\right\}
$$

Conditions (8.8) are the necessary conditions for an extremum. The sufficient conditions for an extremum for the function $z=f(x, y)$ are expressed using the determinant

$$
\Delta=\left|\begin{array}{ll}
A & B \\
B & C
\end{array}\right|=A C-B^{2}
$$

where

$$
A=f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right), B=f_{x y}^{\prime \prime}\left(x_{0}, y_{0}\right), C=f_{y y}^{\prime \prime}\left(x_{0}, y_{0}\right)
$$

## a. have the following:

1) if $\Delta>0$, then $M_{0}\left(x_{0}, y_{0}\right)$ is an extremum point: if $A>0$ (or $C>0$) - a point of maximum, if $A<0$ (or $C<0$) - a point of minimum,
2) if $\Delta<0$, then there is no extremum at the point $M_{0}$.

If $\Delta=0$, the question of the presence or absence of an extremum of the function remains open (further investigation of the function is required, for example, by the sign of the increment $\Delta f$ near this point).