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I. Differentiation – 7. Lagrangemetoden 144 y 1 f(x,y)=0 f(x,y) = x + 3y, g(x,y) = x 2 + y 2 = 10 7.6. Maksimum/minimum under bibetingelse ☞ [S] 11.8 Lagrange multipliers Eksempel - fortsat Niveaukurverne x + 3y = c tangerer cirklen x 2 + y 2 = 10 for værdier af c, hvor har dobbeltrod. Diskriminanten (−3y + c) 2 + y 2 − 10 = 10y 2 − 6cy + c 2 − 10 36c 2 − 40(c 2 − 10) = −4c 2 + 400 x
I. Differentiation – 7. Lagrangemetoden 145 er 0 for c = ±10, som giver punkter (x,y) = ±(1,3) 7.7. Lagrange multiplikator ☞ [S] 11.8 Lagrange multipliers Definition Bestem ekstremumspunkter for en funktion f(x,y) under begræsningen g(x,y) = k. I ligningen ∇f(x0,y0) = λ∇g(x0,y0) kaldes den ubekendte λ for en Lagrange multiplikator. Ligningen udtrykker at niveaukurven for f i (x0,y0) tangerer begræsningskurven g(x,y) = k. 7.8. Lagrange multiplikator metode ☞ [S] 11.8 Lagrange multipliers Metode Bestem ekstremumspunkter for en funktion f(x,y) under begræsningen g(x,y) = k. (a) Find x,y,λ så ∇f(x,y) = λ∇g(x,y) g(x,y) = k (b) Bestem funktionsværdierne i punkterne fra (a). Maksimum og minimum er blandt disse.
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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V. Matricer - 4. Determinanter = (
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Litteratur • Arnold, Ordinary dif
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