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Web skærmformat. - Aarhus Universitet

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I. Differentiation – 6. Maksimum/minimum 127<br />

z<br />

har et saddelpunkt i (0,0).<br />

x<br />

z = 1 − x 2 + y 2<br />

6.25. Ekstremumspunkters type ☞ [S] 11.7 Maximum and minimum values<br />

Eksempel 3<br />

f(x,y) = x 4 + y 4 − 4xy + 1<br />

har kritiske punkter, hvor<br />

De kritiske punkter bestemmes<br />

∇f(x,y) = (4x 3 − 4y,4y 3 − 4x) = (0,0)<br />

x 3 − y = 0, y 3 − x = 0 ⇔<br />

x 3 − y = 0, (x 3 ) 3 − x = 0 ⇔<br />

y

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