Predavanje 11.5.
Primeri. $$ \begin{aligned} \max / \min &\ 2x^3 + 4x^2 + y^2 - 2xy \\ x &\in \Omega = \mathbb{R}^2 \\ y - 4 &\le 0 \\ x^2 - y &\le 0 \end{aligned} $$
način:
- V notranjosti: $$ \begin{alignedat}{2} {\partial f \over \partial x}(x, y) &=&\ 6x^2 + 8x - 2y &= 0 \\ {\partial f \over \partial y}(x, y) &=&\ 2y - 2x &= 0 \Longrightarrow x = y \\ && 6x^2 + 6x &= 0 \\ && x(x+1) &= 0 \end{alignedat} $$ Dobimo kandidata $x = y = 0$ in $x = y = -1$, ki pa nista v $\mathring{D}$ (relativna notranjost $D$).
- Na zgornjem robu: $x \in (-2, 2)$, $y = 4$ $$ \begin{aligned} g(x) &= f(x, 4) = 2x^3 + 4x^2 - 8x + 16 \\ g'(x) &= 6x^2 + 8x - 8 = 2(3x^2 + 4x - 4) = 0 \\ x &= {-4 \pm \sqrt{16 + 48} \over 6} = {-2 \pm 4 \over 3} \end{aligned} \\ \begin{aligned} x_1 &= -2 & f(-2, 4) &= 32 \quad \text{(ni v relativni notranjosti roba)} \\ x_2 &= {2 \over 3} & f\left({2 \over 3}, 4\right) &= {352 \over 27} \approx 13.037 \end{aligned} $$
- Na spodnjem robu: $x \in (-2, 2)$, $y = x^2$ $$ \begin{aligned} h(x) &= f(x, x^2) = x^4 + 4x^2 \\ h'(x) &= 4x^3 + 8x = 4x(x^2 + 2) = 0 \\ x &= 0 \quad \ \ f(0, 0) = 0 \end{aligned} $$
- Na stičiščih obeh robov: $x = \pm 2$, $y = 4$ $$ f(-2, 4) = 32 \qquad f(2, 4) = 32 $$ Globalni maksimum je torej v $(\pm 2, 4)$, globalni minimum pa v $(0, 0)$.
način:
- V notranjosti: kot prej.
- Na zgornjem robu: $y - 4 = 0$. Definirajmo Lagrangeevo funkcijo $$ L(x, y, \lambda) = 2x^3 + 4x^2 + y^2 - 2xy + \lambda (y - 4), $$ kjer je $\lambda$ Lagrangeev množitelj (multiplikator). Poiščimo parcialne odvode Lagrangeeve funkcije: $$ \begin{alignedat}{2} {\partial L \over \partial x} &= 6x^2 + 8x - 2y &&= 0 \\ {\partial L \over \partial y} &= 2y - 2x + \lambda &&= 0 \\ {\partial L \over \partial \lambda} &= y - 4 &&= 0 \end{alignedat} \\ y = 4 \qquad x = {-2 \pm 4 \over 3} \qquad \lambda = {-28 \pm 8 \over 3} $$ Sistem enačb ima rešitvi $(x, y, \lambda) = (-2, 4, -12)$ in $(x, y, \lambda) = \left({2 \over 3}, 4, -{20 \over 3}\right)$.
- Na spodnjem robu: $$ L(x, y, \mu) = 2x^3 + 4x^2 + y^2 - 2xy + \mu (x^2 - y) \\ \begin{alignedat}{2} {\partial L \over \partial x} &= 6x^2 + 8x - 2y + 2 \mu x &&= 0 \\ {\partial L \over \partial y} &= 2y - 2x - \mu &&= 0 \\ {\partial L \over \partial \mu} &= x^2 - y &&= 0 \\[2ex] y &= x^2 \\ \mu &= 2x^2 - 2x \\ 0 &= 4x^3 + 8x = 4x (x^2 + 2) \end{alignedat} \\ x = 0 \qquad y = 0 \qquad \mu = 0 $$ Sistem enačb ima rešitev $(x, y, \mu) = (0, 0, 0)$.
- Na stičišču obeh robov: $$ L(x, y, \lambda, \mu) = 2x^3 + 4x^2 + y^2 - 2xy + \lambda (y - 4) + \mu (x^2 - y) \\ \begin{alignedat}{2} {\partial L \over \partial x} &= 6x^2 + 8x - 2y + 2 \mu x &&= 0 \\ {\partial L \over \partial y} &= 2y - 2x + \lambda - \mu &&= 0 \\ {\partial L \over \partial \lambda} &= y - 4 &&= 0 \\ {\partial L \over \partial \mu} &= x^2 - y &&= 0 \\[2ex] \end{alignedat} \\ y = 4 \qquad x = \pm 2 \qquad \mu = \mp 4 - 4 \qquad \lambda = -12 $$ Sistem enačb ima rešitvi $(x, y, \lambda, \mu) = (4, -2, -12, 0)$ in $(x, y, \lambda, \mu) = (4, 2, -12, -8)$.
način: $$ \begin{alignedat}{2} f(x, y) &= 2x^3 + 4x^2 + y^2 - 2xy \\ \text{p.p.} \quad y - 4 &\le 0 \\ x - y^2 &\le 0 \\[2ex] L(x, y, \lambda, \mu) &= 2x^3 + 4x^2 + y^2 - 2xy &&+ \lambda (y - 4) + \mu (x^2 - y) \\ L_x := {\partial L \over \partial x} &= 6x^2 + 8x - 2y + 2 \mu x &&= 0 \\ L_y := {\partial L \over \partial y} &= 2y - 2x + \lambda - \mu &&= 0 \\ \end{alignedat} $$ Ločimo primere:
- $\lambda = 0$, $\mu = 0$: notranjost
- $y - 4 = 0$, $\mu = 0$: zgornji rob
- $\lambda = 0$, $x^2 - y = 0$: spodnji rob
- $y - 4 = 0$, $x^2 - y = 0$: stičišči robov
Za problem vezanih ekstremov z neenačbami lahko torej zapišemo Lagrangeevo funkcijo $$ L(x_1, \dots, x_n, \lambda_1, \dots, \lambda_m) = f(x_1, \dots, x_n) + \sum_{i=1}^m \lambda_i g_i(x_1, \dots, x_n) $$ Lokalni ekstremi so doseženi pri $$ \begin{align} L_{x_j} := {\partial L \over \partial x_j}(x_1, \dots, x_n, \lambda_1, \dots, \lambda_m) &= 0 & (1 \le j \le n) \\ \lambda_i g_i(x_1, \dots, x_n) &= 0 & (1 \le i \le m) \\ g_i(x_1, \dots, x_n) &\le 0 & (1 \le i \le m) \end{align} $$ Ločimo na $2^m$ primerov glede na $g_i(x_1, \dots, x_n) = 0$ oziroma $g_i(x_1, \dots, x_n) < 0$ in posledično $\lambda_i = 0$ ($1 \le i \le m$).