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I. Differentiation – 7. Lagrangemetoden 160 (x,y,z) = ( 2 √ 29 , − 5 √ 29 ,1 − 7 √ 29 ) Ved indsættelse i f = x + 2y + 3z fås 3 + √ 29 og 3 − √ 29. Det første punkt er maksimum og det andet punkt er minimum. 7.35. Opgave ☞ Matematik Alfa 1, August 2000 Opgave 7 Minimer funktionen f(x,y,z) = x 2 + y 2 + z 2 under bibetingelsen 2x + y + z = 1. Løsning f angiver kvadratet for afstanden fra 0 til planen givet ved bibetingelsen. Fra lineær algebra ved vi at minimum antages. Kandidater til minimumspunkt findes ved Lagrange metoden. 7.36. Opgave ☞ Matematik Alfa 1, August 2000 Opgave 7 - fortsat Lagrange metoden anvendes på funktionen f(x,y,z) = x 2 +y 2 +z 2 under bibetingelsen g(x,y,z) = 2x + y + z = 1. Løsning De partielle afledede er fx = 2x,fy = 2y,fz = 2z gx = 2,gy = 1,gz = 1
I. Differentiation – 7. Lagrangemetoden 161 7.37. Opgave ☞ Matematik Alfa 1, August 2000 Opgave 7 - fortsat Lagrange ligningssystem bliver 2x = 2λ 2y = λ 2z = λ 2x + y + z = 1 Løsningen giver multiplikator λ = 1 3 og minimumspunkt/værdi (a,b,c) = ( 1 3 , 1 6 1 1 , ), f(a,b,c) = 6 6 7.38. Opgave ☞ Matematik Alfa 1, August 2000 Opgave 7 - alternativt Løs bibetingelsen, z = 1 − 2x − y og minimer funktionen h(x,y) = f(x,y,1 − 2x − y) = x 2 + y 2 + (1 − 2x − y) 2 = 5x 2 + 2y 2 + 4xy − 4x − 2y + 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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