MATHHX A

MATHHX A

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4.6 Ekstrema for funktioner af to variable

I dette afsnit skal vi supplere vores differentialregningsværktøjskasse (sejt ord), så vi er rustet til de udfordringer vi møder i næste afsnit. Afsnittet har i sig selv ikke noget med lineær regression at gøre, men det gør det jo kun mere interessant.

Partielle afledede

Ligesom man kan differentiere funktioner af én variable, kan man også differentiere funktioner af to variable. Det gør man ved at betragte den ene variabel som en konstant. På den måde får man en funktion af én variabel, som man kan differentiere på sædvanlig vis. Det er nemmest at forklare ved et eksempel.

  • Eksempel 4.6.1
    Lad \(f(x,y)=y^2+x\). Vi vil nu finde det som kaldes den partielle afledede med hensyn til y og betegnes \(\frac {\partial }{\partial y}f(x,y)\). Det gør vi ved at opfatte \(x\) som en konstant, og differentiere \(f\) som funktion af \(y\). Vi ved at \(y^2\) bliver \(2y\) når man differentierer, og \(x\) skulle betragtes som en konstant, så den bliver til \(0\). Altså får vi

    \[\frac {\partial }{\partial y}f(x,y)=2y+0=2y\]

    Vi kan også finde den partielle afledede med hensyn til x. Det gøres selvfølgelig ved at holde \(y\) konstant, og differentiere funktionen som en funktion af \(x\). Hvis \(y\) er en konstant er \(y^2\) også en konstant, og den bliver derfor \(0\) når vi differentiere. Når vi differentiere \(x\) får vi \(1\) og altså har vi:

    \[\frac {\partial }{\partial x}f(x,y)=0+1=1.\]

Øvelse 4.6.1

Bestem både \(\frac {\partial }{\partial x}f(x,y)\) og \(\frac {\partial }{\partial y}f(x,y)\) for følgende funktioner

  • a) \(f(x,y)=x-y+2\)

  • b) \(f(x,y)=x^3y^2\)

Løsning 4.6.1

  • a) \(\frac {\partial }{\partial x}f(x,y)=1\) og \(\frac {\partial }{\partial y}f(x,y)=-1\)

  • b) \(\frac {\partial }{\partial x}f(x,y)=3x^2y^2\) og \(\frac {\partial }{\partial y}f(x,y)=2x^3y\)

Det er ikke altid at funktions variable hedder \(x\) og \(y\).

  • Eksempel 4.6.2
    Lad \(f(v,w)=2w^2-\ln (v)+0\). Vi kan se at \(f\) en funktion af \(v\) og \(w\). Vi finder:

    \[\frac {\partial }{\partial v}f(v,w)=0-\frac {1}{v}-0=-\frac {1}{v}\]

    og

    \[\frac {\partial }{\partial w}f(v,w)=2\cdot 2w-0+0=4w\]

Øvelse 4.6.2

Bestem:

  • a) \(\frac {\partial }{\partial a}f(a,b)\) og \(\frac {\partial }{\partial b}f(a,b)\) for \(f(a,b)=2a+ab^2-4\)

  • b) \(\frac {\partial }{\partial p}f(p,q)\) og \(\frac {\partial }{\partial q}f(p,q)\) for \(f(p,q)=e^q+b^2-3p\)

Løsning 4.6.2

De partielle afledede er:

  • a) \(\frac {\partial }{\partial a}f(a,b)=2+b^2\) og \(\frac {\partial f}{\partial b}f(p,q)=2ab\)

  • b) \(\frac {\partial }{\partial p}f(p,q)=-3\) og \(\frac {\partial }{\partial q}f(p,q)=e^q\)

Stationære punkter

Vi husker følgende sætning for funktioner af én variabel:

  • Sætning  (Fra Mat-B)
    Lad \(f\) være en differentialbel funktion som er defineret på et åbent interval \(I\).

    Hvis \(f\) har et ekstremum i \(x_0\in I\), så er \(f'(x_0)=0\).

Der gælder noget tilsvarende for funktionerne af to variable:

  • Sætning 4.6.1
    Lad \(f(x,y)\) være en funktion af to variable defineret på en åben mængde \(O\) og antag at begge de partielle afledede \(\frac {\partial }{\partial x}f(x,y)\) og \(\frac {\partial }{\partial y}f(x,y)\) eksisterer.

    Hvis \(f\) har et ekstremum i \((x_0,y_0)\in O\), så er \(\frac {\partial }{\partial x}f(x_0,y_0)=0\) og \(\frac {\partial }{\partial y}f(x_0,y_0)=0\).

Hvad en ”åben mængde” betyder i denne sammenhæng, vil vi ikke gå i detaljer med, vi vil blot hæfte os ved, at hvis vi skal finde ekstrema for en funktion af to variable, så skal vi kigge efter de punkter hvor begge partielle afledede er nul. Et sådan punkt kaldes et stationært punkt. Vi ved altså at evt. ekstrema vil ligge i de stationære punkter. Vi kan dog ikke være sikker på, at bare fordi en funktion har et stationært punkt, så har den også ekstremum i punktet. Det er ligesom med \(f(x)=x^3\). Her er \(f'(0)=0\), men \(f\) har ikke ekstremum i \(x=0\).

  • Eksempel 4.6.3
    Vi vil bestemme de stationære punkter for funktionen

    \[f(x,y)=x^2 +2y^2-xy+14x\]

    Vi bestemmer først de to partielle afledede:

    \[\frac {\partial }{\partial x}f(x,y)=2x-y+14\]

    og

    \[\frac {\partial }{\partial y}f(x,y)=4y-x\]

    De partielle afledede skal begge være nul så:

    \[2x-y+14=0\quad \textrm {og}\quad 4y-x=0\]

    Vi isolerer \(x\) i den anden ligning:

    \[x=4y\]

    og sætter værdien for \(x\) ind i den første

    \begin{align*} 2\cdot (4y)-y+14&=0 \\ 7y +14 & = 0\\ y&=-2 \end{align*} Vi indsætter nu værdien for \(y\) i den anden ligning \(4y-x=0\)

    \begin{align*} 4\cdot (-2)-x&=0 \\ x & = -8\\ \end{align*} Vi konkluderer at \(f\) har et stationærtpunkt i \((-8,-2)\)

Øvelse 4.6.3

Bestem stationære punkter for følgende funktioner.

  • a) \(f(x,y) = x^2 + 2y^2 - 2xy + 10y\)

  • b) \(f(x,y) =xy-y\)

  • c) \(f(x,y)= x^2 - y^2 - 2y\)

Løsning 4.6.3

  • a) \((-5,-5)\)

  • b) \((1,0)\)

  • c) \((0,-1)\)

Øvelse 4.6.4

Betragt funktionen \(f(a,b)= (2a-b-1)^2+(a-2b+13)^2\)

  • a) Regn de partielle aflede \(\frac {\partial }{\partial a}f(a,b)\) og \(\frac {\partial }{\partial b}f(a,b)\). I stedet for at hæve parenteserne skal du bruge reglen om differentiation af sammensatte funktioner.

  • b) Bestem de stationære punkter for \(f\).

Løsning 4.6.4

  • a) \(\frac {\partial }{\partial a}f(a,b)=10a-8b+22\) og \(\frac {\partial }{\partial b}f(a,b)=-8a+10b-50\).

  • b) \((5,9)\).

Det store spørgsmål er nu, hvordan man afgør om et stationært punkt er et ekstremum og i givet fald, hvilket slags ekstremum. Det er selvfølgeligt vældigt interessant, men i forhold til lineær regression er det nok for os, at kunne identificere de stationære punkter.