Investigation of functions for local extrema.

An extremum refers to a maximum or minimum of a function.

Necessary condition for an extremum.

If the differentiable function $f (P)$ reaches an extremum at point $P_{0}$ , then at this point

$f_{x_{k}} (P_{0}) = 0$ for all $k = 1, 2, . . ., n (1)$

Points where conditions (1) are satisfied are called stationary points of the function $u = f (P)$ .

Sufficient conditions for extrema of a function of two variables.

Let $P_{0} (x_{0}, y_{0})$ be a stationary point of the function $z = f (x, y)$ , and this function is twice differentiable in some neighborhood of the point, and all its second partial derivatives are continuous at the point.

Let $Δ$ denote the determinant

$Δ (x; y) = | \begin{matrix} z_{x x}^{″} & z_{x y}^{″} \\ z_{y x}^{″} & z_{y y}^{″} \end{matrix} | = z_{x x}^{″} z_{y y}^{″} - z_{x y}^{″} z_{y x}^{″} = z_{x x}^{″} z_{y y}^{″} - {z^{″}}_{x y}^{2} .$

Then:

a) If $Δ (P_{0}) > 0$ , then there is an extremum at point $P_{0}$ , and in the case $z^{″} x x (P_{0}) > 0$ (or $z^{″} y y (P_{0}) > 0$ ) - it's a minimum, and in the case $z^{″} x x (P_{0}) < 0$ (or $z^{″} y y (P_{0}) < 0$ ) - it's a maximum.

b) If $Δ (P_{0}) < 0$ , then there is no extremum at point $P_{0}$ (such points are called saddle points).

c) If $Δ (P_{0}) = 0$ , then additional investigation using differentials of the third or higher order is necessary to determine the existence of an extremum (indefinite case).

Examples.

Find the extrema of functions of two variables:

7.187. $z = x^{2} + x y + y^{2} - 3 x - 6 y .$

Solution.

Let's find the first-order partial derivatives of the function $z (x, y) :$

$z_{x}^{'} (x, y) = (x^{2} + x y + y^{2} - 3 x - 6 y)_{x}^{'} = 2 x + y - 3;$

$z_{y}^{'} (x, y) = (x^{2} + x y + y^{2} - 3 x - 6 y)_{y}^{'} = x + 2 y - 6.$

Next, we find the stationary points by setting the found derivatives to zero and solving the system:

${\begin{array}{lcl} 2 x + y - 3 = 0 \\ x + 2 y - 6 = 0 \end{array} \Rightarrow {\begin{array}{lcl} y = 3 - 2 x \\ x + 2 (3 - 2 x) - 6 = 0 \end{array} \Rightarrow {\begin{array}{lcl} y = 3 - 2 x \\ - 3 x = 0 \end{array} \Rightarrow {\begin{array}{lcl} y = 3 \\ x = 0 \end{array}$

Thus, we have found the stationary point $P (0; 3)$ . Let's check if it is an extremum point. To do this, we find the second derivatives:

$z_{x x}^{″} (x, y) = (2 x + y - 3)_{x}^{'} = 2;$

$z_{y y}^{″} (x, y) = (x + 2 y - 6)_{y}^{'} = 2.$

$z_{x y}^{″} (x, y) = (2 x + y - 3)_{y}^{'} = 1.$

Next, we find

$Δ (x; y) = | \begin{matrix} z_{x x}^{″} & z_{x y}^{″} \\ z_{y x}^{″} & z_{y y}^{″} \end{matrix} | = | \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} | = 4 - 1 = 3.$

Since $Δ (P_{0}) > 0$ , the point $P_{0}$ is an extremum point, and since $z_{x x}^{″} (P_{0}) > 0$ , it is a minimum.

$z_{m i n} = 0^{2} + 0 + 3^{2} - 0 - 18 = - 9.$

Answer: The function has a minimum at point $P (0; 3)$ . $z_{m i n} = - 9.$

7.189. $z = 3 x^{2} - x^{3} + 3 y^{2} + 4 y .$

Solution.

Let's find the first-order partial derivatives of the function $z (x, y) :$

$z_{x}^{'} (x, y) = (3 x^{2} - x^{3} + 3 y^{2} + 4 y)_{x}^{'} = 6 x - 3 x^{2};$

$z_{y}^{'} (x, y) = (3 x^{2} - x^{3} + 3 y^{2} + 4 y)_{x}^{'} = 6 y + 4;$

Next, we find the stationary points by setting the found derivatives to zero and solving the system:

${\begin{array}{lcl} 6 x - 3 x^{2} = 0 \\ 6 y + 4 = 0 \end{array} \Rightarrow {\begin{array}{lcl} [\begin{array}{lcl} x = 0 \\ 6 - 3 x = 0 \end{array} \\ y = - \frac{2}{3} \end{array} \Rightarrow {\begin{array}{lcl} [\begin{array}{lcl} x = 0 \\ x = 2 \end{array} \\ y = - \frac{2}{3} \end{array}$

Thus, we have two stationary points $P_{1} (0; - \frac{2}{3})$ and $P_{2} (2; - \frac{2}{3})$ . Let's check if they are extremum points. To do this, we find the second derivatives:

$z_{x x}^{″} (x, y) = (6 x - 3 x^{2})_{x}^{'} = 6 - 6 x;$

$z_{y y}^{″} (x, y) = (6 y + 4)_{y}^{'} = 6.$

$z_{x y}^{″} (x, y) = (6 x - 3 x^{2})_{y}^{'} = 0.$

Next, we find

$Δ (x; y) = | \begin{matrix} z_{x x}^{″} & z_{x y}^{″} \\ z_{y x}^{″} & z_{y y}^{″} \end{matrix} | = | \begin{matrix} 6 - 6 x & 0 \\ 0 & 6 \end{matrix} | = 36 - 36 x .$

$Δ (0; - \frac{2}{3}) = 36.$

$Δ (2; - \frac{2}{3}) = 36 - 72 = - 36.$

Since $Δ (P_{1}) > 0$ , the point $P_{1}$ is an extremum point, and since $z_{x x}^{″} (P_{1}) = 6 > 0$ , it is a minimum.

$z_{m i n} = 0 + 3 {(- \frac{2}{3})}^{2} - 4 \frac{\cdot}{2} 3 = \frac{4}{3} - \frac{8}{3} = - \frac{4}{3} .$

Since $Δ (P_{2}) < 0$ , there is no extremum at point $P_{2}$ .

Answer: The function has a minimum at point $P_{1} (0; - \frac{2}{3})$ $z_{m i n} = - \frac{4}{3}$ .

7.193. $z = 2 x^{3} - x y^{2} + 5 x^{2} + y^{2} .$

Solution.

Let's find the first-order partial derivatives of the function $z (x, y) :$

$z_{x}^{'} (x, y) = (2 x^{3} - x y^{2} + 5 x^{2} + y^{2})_{x}^{'} = 6 x^{2} - y^{2} + 10 x;$

$z_{y}^{'} (x, y) = (2 x^{3} - x y^{2} + 5 x^{2} + y^{2})_{y}^{'} = - 2 x y + 2 y .$

Next, we find the stationary points by setting the found derivatives to zero and solving the system:

${\begin{array}{lcl} 6 x^{2} - y^{2} + 10 x = 0 \\ - 2 x y + 2 y = 0 \end{array} \Rightarrow {\begin{array}{lcl} 6 x^{2} - y^{2} + 10 x = 0 \\ [\begin{array}{lcl} y = 0 \\ x = 1 \end{array} \end{array} \Rightarrow [\begin{array}{lcl} {\begin{array}{lcl} y = 0 \\ 6 x^{2} + 10 x = 0 \end{array} \\ {\begin{array}{lcl} x = 1 \\ 6 - y^{2} + 10 = 0 \end{array} \end{array}$

From here, we find four stationary points: $P_{1} (- \frac{5}{3}; 0),$ $P_{2} (0; 0),$ $P_{3} (1; 4),$ $P_{4} (1; - 4)$ . Let's check if they are extremum points. To do this, we find the second derivatives:

$z_{x x}^{″} (x, y) = (6 x^{2} - y^{2} + 10 x)_{x}^{'} = 12 x + 10;$

$z_{y y}^{″} (x, y) = (- 2 x y + 2 y)_{y}^{'} = - 2 x + 2.$

$z_{x y}^{″} (x, y) = (6 x^{2} - y^{2} + 10 x)_{y}^{'} = - 2 y .$

Next, we find

$Δ (x; y) = | \begin{matrix} z_{x x}^{″} & z_{x y}^{″} \\ z_{y x}^{″} & z_{y y}^{″} \end{matrix} | = | \begin{matrix} 12 x + 10 & - 2 y \\ - 2 y & - 2 x + 2 \end{matrix} | = (12 x + 10) (- 2 x + 2) - 4 y^{2} =$ $= - 24 x^{2} + 4 x + 20 - 4 y^{2}$