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I. Differentiation – 6. Maksimum/minimum 130
Foruden (x,y) = (0,0) fås, x = 0
10y − 5 − 2x 2 = 0, 5x 2 − 4y − 4y 3 = 0 ⇔
10y − 5 − 2x 2 = 0, 8y 3 − 42y + 25 = 0 ⇔
(x,y) ≈ (0,0),(±2.64,1.90),(±0.86,0.65)
... ⇔
6.30. Konklusion ☞ [S] 11.7 Maximum and minimum values
Eksempel 4 - fortsat
fx = 20xy − 10x − 4x 3 , fy = 10x 2 − 8y − 8y 3
fxx = 20y − 10 − 12x 2 , fxy = 20x, fyy = 8 − 24y 2
(a,b) f(a,b) fxx(a,b) D(a,b) Type
(0,0) 0.00 −10.00 80.00 maksimum
(−2.64,1.90) 8.5 −55.93 2488.71 maksimum
(2.64,1.90) 8.5 −55.93 2488.71 maksimum
(−0.86,0.65) −1.48 −5.87 −187.64 saddel
(0.86,0.65) −1.48 −5.87 −187.64 saddel
6.31. Kassefabrikant ☞ [S] 11.7 Maximum and minimum values
Eksempel 6