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I. Differentiation – 7. Lagrangemetoden 150 Eksempel 1 - fortsat Lagranges ligningssystem opskrives Heraf yz = λ(y + 2z) xz = λ(x + 2z) 2xz + 2yz + xy = 12 2xz + 2yz + xy = 12 xy = λ(2x + 2y) xyz = λ(xy + 2xz) xyz = λ(xy + 2yz) xyz = λ(2xz + 2yz) 7.18. Kassefabrikant ☞ [S] 11.8 Lagrange multipliers Eksempel 1 - fortsat Der løses x = y y = 2z 4z 2 + 4z 2 + 4z 2 = 12
I. Differentiation – 7. Lagrangemetoden 151 Løsningen er da Kantlængder for størst rumfang er (x,y,z) = (2,2,1) længde = 2m, bredde = 2m, højde = 1m 7.19. Køreplan ☞ [S] 11.7 Maximum and minimum values 9 Bemærkning Find absolut maksimum og minimum for en kontinuert funktion f på en lukket og begrænset mængde D: 1. Find værdier af f i kritiske punkter i D 2. Find ekstremværdier af f på randen af D 3. Vælg maksimum/minimum fra 1. og 2. 7.20. Find ekstremumspunkter ☞ [S] 11.8 Lagrange multipliers Eksempel 2, 3 Bestem ekstremumsværdier af f(x,y) = x 2 + 2y 2 på cirkelskiven Løsning D = {(x,y)|x 2 + y 2 ≤ 1}
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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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1. Grafiske/numeriske metoder IV Di
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1. Vektorer og matricer V Matricer
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V. Matricer - 4. Determinanter 419
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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Identitetsmatricen, 383 Identitetsm
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