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I. Differentiation – 7. Lagrangemetoden 158 Ligningen udtrykker at niveaufladen for f i (x0,y0,z0) tangerer begræsningskurven g(x,y,z) = k, h(x,y,z) = c. 7.31. Lagrange multiplikator metode ☞ [S] 11.8 Lagrange multipliers Metode Bestem ekstremumspunkter for en funktion f(x,y,z) under begræsningen g(x,y,z) = k, h(x,y,z) = c. (a) Find x,y,z,λ,µ så ∇f(x,y,z) = λ∇g(x,y,z) + µ∇h(x,y,z) g(x,y,z) = k h(x,y,z) = c (b) Bestem funktionsværdierne i punkterne fra (a). Maksimum og minimum er blandt disse. 7.32. Lagrange multiplikator metode ☞ [S] 11.8 Lagrange multipliers Ligninger Lagranges ligningssystem for bestemmelse af ekstremumspunkter for en funktion f(x,y,z) under begræsningen g(x,y,z) = k, h(x,y,z) = c. fx(x,y,z) = λgx(x,y,z) + µhx(x,y,z) fy(x,y,z) = λgy(x,y,z) + µhy(x,y,z) fz(x,y,z) = λgz(x,y,z) + µhz(x,y,z)
I. Differentiation – 7. Lagrangemetoden 159 g(x,y,z) = k h(x,y,z) = c 7.33. To betingelser ☞ [S] 11.8 Lagrange multipliers Eksempel 5 Lagranges ligningssystem for bestemmelse af ekstremumspunkter for funktion f = x + 2y + 3z under begræsningen g = x − y + z = 1, h = x 2 + y 2 = 1. 1 = λ + µ2x 2 = −λ + µ2y 3 = λ x − y + z = 1 x 2 + y 2 = 1 7.34. To betingelser ☞ [S] 11.8 Lagrange multipliers Eksempel 5 - fortsat Løsningen giver relevante punkter (x,y,z) = (− 2 √ 29 , 5 √ 29 ,1 + 7 √ 29 )
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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