402 lines
11 KiB
Plaintext
402 lines
11 KiB
Plaintext
|
{
|
||
|
|
"cells": [
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "4cc6c96f",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"# Ćwiczenia 9"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "1276423e",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"***TEMAT:*** funkcje dwóch zmiennych - ekstrema lokalne"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "bc2b4d7c",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"## Ekstrema lokalne"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "9fdf1c56",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Funkcja $f$ (dwóch zmiennych) określona na obszarze otwartym $D$ ma w punkcie $(x_0,y_0)\\in D$ maksimum lokalne, jeśli istnieje otoczenie $U$ tego punktu, takie że w każdym punkcie $(x,y)\\in U$ różnym od $(x_0,y_0)$, zachodzi nierówność $f(x_0,y_0)\\ge f(x,y)$ (dla minimum $\\le$). Jeśli nierówność jest ostra na pewnym otoczeniu różnym od $(x_0,y_0)$, to ekstremum nazywa się _ścisłe_."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "042d333c-ecb1-4b4d-b170-9343b7c8ddcb",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"### Warunek konieczny istnienia ekstremum"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "c430e78c",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"__Twierdzenie__ (warunek konieczny istnienia ekstremum)\n",
|
||
|
|
"\n",
|
||
|
|
" Jeśli $f$ ma w $(x_0,y_0)$ ekstremum i ma tam obie pochodne cząstkowe, to obie są równe zeru. Inaczej mówiąc:\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\begin{cases}\n",
|
||
|
|
"\\frac {\\partial f}{\\partial x}(x_0,y_0)=0;\\\\\n",
|
||
|
|
"\\frac {\\partial f}{\\partial y}(x_0,y_0)=0.\n",
|
||
|
|
"\\end{cases}\n",
|
||
|
|
"$$"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "f21ec660",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"#### Zadanie 1 (punkt siodłowy)\n",
|
||
|
|
"\n",
|
||
|
|
"W których punktach może mieć ekstremum funkcja $f(x,y)=x^2-y^2$?"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "ae83f501-f3f8-428d-a27e-ffec022316a8",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"##### Rozwiązanie:"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "ad2937e7-1541-4c36-b852-29b894489cf5",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Jedynym rozwiązaniem układu równań\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\begin{cases}\n",
|
||
|
|
"2x=0;\\\\\n",
|
||
|
|
"2y=0;\n",
|
||
|
|
"\\end{cases}\n",
|
||
|
|
"$$\n",
|
||
|
|
"jest punkt $A=(0,0)$.\n",
|
||
|
|
"\n",
|
||
|
|
"Ponieważ \n",
|
||
|
|
"\n",
|
||
|
|
"$$f(x,0)>0 \\text{ dla wszystkich }x\\not=0$$\n",
|
||
|
|
"oraz \n",
|
||
|
|
"$$f(0,y)<0 \\text{ dla wszystkich }y\\not=0,$$\n",
|
||
|
|
"to punkt $(0,0)$ nie jest ekstremum lokalnym."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "74de856d-acb6-4495-93dc-4efc27aacd13",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"#### Zadanie 2 \n",
|
||
|
|
"Czy funkcja $f(x,y)=|x|+|y|$\n",
|
||
|
|
"ma w punkcie $(0,0)$ pochodne cząstkowe? Czy ma w tym punkcie ekstremum lokalne?"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "eedd4774-ceba-43bd-8dbe-5aa9e191a14d",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"##### Rozwiązanie:\n",
|
||
|
|
"\n",
|
||
|
|
"Ponieważ $f(0,0)=0$ oraz $f(x,y)>0$ dla $(x,y)\\not=(0,0)$, to w punkcie $(0,0)$ \n",
|
||
|
|
"funkcja $f$ ma silne minimum lokalne. \n",
|
||
|
|
"Pochodne cząstkowe funkcji $f$ nie istnieją w $(0,0)$, bo granica\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\lim_{h\\to 0}\\frac{|h|}{h}\n",
|
||
|
|
"$$\n",
|
||
|
|
"nie istnieje."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "5177becf-f4cc-4564-bec0-43444a32e369",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"### Warunek dostateczny istnienia ekstremum"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "3384c139",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
" __Twierdzenie__ (warunek wystarczający istnienia ekstremum.)\n",
|
||
|
|
"\n",
|
||
|
|
" Niech funkcja $f$ ma ciągłe pochodne cząstkowe II rzędu na otoczeniu punktu $(x_0,y_0)$ oraz niech:\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\begin{cases}\n",
|
||
|
|
"\\frac {\\partial f}{\\partial x}(x_0,y_0)=0\\\\\n",
|
||
|
|
"\\frac {\\partial f}{\\partial y}(x_0,y_0)=0\n",
|
||
|
|
"\\end{cases}\n",
|
||
|
|
"\\quad \\text{oraz}\\quad \n",
|
||
|
|
"H(x_0,y_0)=\\begin{vmatrix}\n",
|
||
|
|
"\\frac{\\partial^2 f}{\\partial x^2}(x_0,y_0)& \\frac{\\partial^2 f}{\\partial x\\partial y}(x_0,y_0)\\\\\n",
|
||
|
|
"\\frac{\\partial^2 f}{\\partial y\\partial x}(x_0,y_0)& \\frac{\\partial^2 f}{\\partial y^2}(x_0,y_0)\n",
|
||
|
|
"\\end{vmatrix} > 0.\n",
|
||
|
|
"$$\n",
|
||
|
|
"Wtedy w punkcie $(x_0,y_0)$ funkcja ma ekstremum lokalne właściwe. Jest to maksimum, gdy którakolwiek pochodna cząstkowa II rzędu nie-mieszana jest ujemna, minimum - gdy dodatnia. \n",
|
||
|
|
"\n",
|
||
|
|
"Jeżeli $H(x_0,y_0)<0$, to ekstremum nie ma, a jeżeli $H(x_0,y_0)=0$, to nie wiadomo."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "88e2c6b0-148a-4a39-b5a7-af63a40aa871",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"#### Zadanie 3\n",
|
||
|
|
"\n",
|
||
|
|
"Wyznaczymy ekstrema lokalne dla funkcji \n",
|
||
|
|
"$$ \n",
|
||
|
|
"f(x,y)=x^3+3xy^2-15 x-12y.\n",
|
||
|
|
"$$"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "08eecc86-a7fa-4869-9e7f-4900c2d9abfc",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"##### Rozwiązanie:"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "491f88ce-f8bb-45f0-a02f-6a017355b097",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Mamy\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\frac{\\partial f}{\\partial x}(x,y)=3x^2+3y^2-15 \\quad\\text{ oraz }\\quad \\frac{\\partial f}{\\partial y}(x,y)=6xy-12.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "4c01d0ae-38dc-46f8-a9d1-903d3c7147b1",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Rozwiązaniami układu równań (wystarczy wyznaczyć $y$ z drugiego równanania i wstawić do pierwszego)\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\begin{cases}\n",
|
||
|
|
"3x^2+3y^2-15=0\\\\\n",
|
||
|
|
"6xy-12=0\n",
|
||
|
|
"\\end{cases}\n",
|
||
|
|
"$$\n",
|
||
|
|
"są punkty $A_1=(1,2)$, $A_2=(-1,-2)$, $A_3=(2,1)$, $A_4=(-2,-1)$."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "b9462877-8cce-4540-808a-43dfbece08b7",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Mamy\n",
|
||
|
|
"$$\n",
|
||
|
|
"H(x_0,y_0)=\\begin{vmatrix}\n",
|
||
|
|
"\\frac{\\partial^2 f}{\\partial x^2}(x_0,y_0)& \\frac{\\partial^2 f}{\\partial x\\partial y}(x_0,y_0)\\\\\n",
|
||
|
|
"\\frac{\\partial^2 f}{\\partial y\\partial x}(x_0,y_0)& \\frac{\\partial^2 f}{\\partial y^2}(x_0,y_0)\n",
|
||
|
|
"\\end{vmatrix} =\n",
|
||
|
|
"\\begin{vmatrix}\n",
|
||
|
|
"6x_0& 6y_0\\\\\n",
|
||
|
|
"6y_0& 6x_0\n",
|
||
|
|
"\\end{vmatrix}=36(x_0^2-y_0^2).\n",
|
||
|
|
"$$"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "2d193dcb-2a40-4a26-b125-257145bcdadf",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"Stąd:\n",
|
||
|
|
"1. $H(1,2)<0$ zatem punkt $(1,2)$ nie jest ekstremum lokalnym.\n",
|
||
|
|
"2. $H(-1,-2)<0$ zatem punkt $(-1,-2)$ nie jest ekstremum lokalnym.\n",
|
||
|
|
"3. $H(2,1)>0$ i $\\frac{\\partial^2 f}{\\partial x^2}(2,1)>0$ - zatem punkt $(2,1)$ jest minimum lokalnym.\n",
|
||
|
|
"4. $H(-2,-1)>0$ i $\\frac{\\partial^2 f}{\\partial x^2}(-2,-1)<0$ - zatem punkt $(2,1)$ jest maksimum lokalnym."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "10ebfae1-7161-496e-bd6e-0afcaef5e403",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"#### More...\n",
|
||
|
|
"\n",
|
||
|
|
"Wiecej przykładów można znaleźć [tutaj](https://alfa.im.pwr.edu.pl/~kajetano/AM2/fun2var/fun2var-6.html)."
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "4df5a853-c8f8-4fb9-92ef-b9bed7f81f2e",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"### Przykładowe zadanie optymalizacyjne"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "3e27d431-15e7-428d-8ec3-6367237b119f",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"#### Zadanie 4 \n",
|
||
|
|
"Jakie powinny być wymiary prostopadłościennego otwartego od góry zbiornika o objętości \n",
|
||
|
|
"1000 litrów, aby na jego wykonanie zużyć jak najmniej materiału?"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "markdown",
|
||
|
|
"id": "07b00b11-8c1c-473a-9e71-63f782b9ce23",
|
||
|
|
"metadata": {},
|
||
|
|
"source": [
|
||
|
|
"##### Rozwiązanie:\n",
|
||
|
|
"Trzeba znaleźć prostopadłościan o objętośći $ 1000 $ litrów tak, aby suma pól powierzchni ścian bocznych oraz jego podstawy była jak najmniejsza.\n",
|
||
|
|
"\n",
|
||
|
|
"Oznaczmy przez $ x $, $ y $ długości krawędzi podstawy naszego prostopadłościanu oraz przez $ h $ jego wysokość. Ponieważ objętość prostopadłościanu ma być równa $1000$, to\n",
|
||
|
|
"$$\n",
|
||
|
|
"x \\cdot y \\cdot h = 1000.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Suma pola podstawy i powierzchni bocznej naszego prostopadłościanu jest równa\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"p(x,y,h)= x \\cdot y + 2 \\cdot (x + y) \\cdot h.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Z warunku $ x \\cdot y \\cdot h = 1000 $ dostajemy, że $ h = \\frac{1000}{x \\cdot y} $. Zatem\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"p(x,y,h)= x \\cdot y + 2 \\cdot (x + y) \\cdot \\frac{1000}{x \\cdot y} = x \\cdot y + \\frac{2000}{x} + \\frac{2000}{y}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Stąd funkcja $p$ jest w istocie funkcją dwóch zmiennych, oznaczmy ją przez $f(x,y)$.\n",
|
||
|
|
"Mamy\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"f(x, y) = x \\cdot y + \\frac{2000}{x} + \\frac{2000}{y}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Za dziedzinę naszej funkcji przyjmujemy zbiór:\n",
|
||
|
|
"$$\n",
|
||
|
|
"D = \\{(x, y) : x > 0, y > 0\\}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Pochodne cząstkowe funkcji $ f $ są równe:\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\frac{\\partial f}{\\partial x}(x,y) = y - \\frac{2000}{x^2},\n",
|
||
|
|
"\\quad\n",
|
||
|
|
"\\frac{\\partial f}{\\partial y}(x,y) = x - \\frac{2000}{y^2}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Musimy rozwiązać układ równań:\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\begin{cases}\n",
|
||
|
|
"y - \\frac{2000}{x^2} = 0, \\\\\n",
|
||
|
|
"x - \\frac{2000}{y^2} = 0.\n",
|
||
|
|
"\\end{cases}\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Wyznaczamy $ y $ z pierwszego równania i podstawiając do drugiego dostajemy, że :\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"y = \\frac{2000}{x^2}\n",
|
||
|
|
"\\quad\\text{oraz}\\quad\n",
|
||
|
|
"x - \\frac{2000}{\\left(\\frac{2000}{x^2}\\right)^2} = 0.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Przekształcając drugie równanie, otrzymujemy:\n",
|
||
|
|
"$$\n",
|
||
|
|
"x - \\frac{x^4}{2000} = 0.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Jedynym dodatnim rozwiązaniem tego równania jest\n",
|
||
|
|
"$$\n",
|
||
|
|
"x = \\sqrt[3]{2000}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"Proste rachunki dają, że \n",
|
||
|
|
"$$\n",
|
||
|
|
"y = \\frac{2000}{x^2} = \\frac{2000}{\\left(\\sqrt[3]{2000}\\right)^2} = \\sqrt[3]{2000}.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Nie jest trudno zauważyc, że w punkcie $( \\sqrt[3]{2000}, \\sqrt[3]{2000})$ funkcja $f$ przyjmuje wartość najmniejszą na zbiorze $D$, wynika to z następujących obserwacji:\n",
|
||
|
|
"\n",
|
||
|
|
"$$\n",
|
||
|
|
"\\lim_{x\\to 0}f(x,y)=\\infty, \\quad \\lim_{y\\to 0}f(x,y)=\\infty, \\quad \\lim_{x\\to \\infty}f(x,y)=\\infty\\quad \\lim_{y\\to \\infty}f(x,y)=\\infty.\n",
|
||
|
|
"$$\n",
|
||
|
|
"\n",
|
||
|
|
"Ostatecznie wymiary zbiornika to: \n",
|
||
|
|
"$$\n",
|
||
|
|
"x=\\sqrt[3]{2000},\\quad y=\\sqrt[3]{2000},\\quad h=\\frac{\\sqrt[3]{2000}}{2}.\n",
|
||
|
|
"$$"
|
||
|
|
]
|
||
|
|
},
|
||
|
|
{
|
||
|
|
"cell_type": "code",
|
||
|
|
"execution_count": null,
|
||
|
|
"id": "86e171af-0c9c-4d1f-97e6-544b49b70a5e",
|
||
|
|
"metadata": {},
|
||
|
|
"outputs": [],
|
||
|
|
"source": []
|
||
|
|
}
|
||
|
|
],
|
||
|
|
"metadata": {
|
||
|
|
"kernelspec": {
|
||
|
|
"display_name": "Python 3 (ipykernel)",
|
||
|
|
"language": "python",
|
||
|
|
"name": "python3"
|
||
|
|
},
|
||
|
|
"language_info": {
|
||
|
|
"codemirror_mode": {
|
||
|
|
"name": "ipython",
|
||
|
|
"version": 3
|
||
|
|
},
|
||
|
|
"file_extension": ".py",
|
||
|
|
"mimetype": "text/x-python",
|
||
|
|
"name": "python",
|
||
|
|
"nbconvert_exporter": "python",
|
||
|
|
"pygments_lexer": "ipython3",
|
||
|
|
"version": "3.9.2"
|
||
|
|
},
|
||
|
|
"toc": {
|
||
|
|
"base_numbering": 1,
|
||
|
|
"nav_menu": {},
|
||
|
|
"number_sections": false,
|
||
|
|
"sideBar": true,
|
||
|
|
"skip_h1_title": false,
|
||
|
|
"title_cell": "Table of Contents",
|
||
|
|
"title_sidebar": "Contents",
|
||
|
|
"toc_cell": false,
|
||
|
|
"toc_position": {
|
||
|
|
"height": "calc(100% - 180px)",
|
||
|
|
"left": "10px",
|
||
|
|
"top": "150px",
|
||
|
|
"width": "165px"
|
||
|
|
},
|
||
|
|
"toc_section_display": true,
|
||
|
|
"toc_window_display": true
|
||
|
|
}
|
||
|
|
},
|
||
|
|
"nbformat": 4,
|
||
|
|
"nbformat_minor": 5
|
||
|
|
}
|