f(x,y)=−(x2−1)2−(x2y−x−1)2 Find the critical point(s)
fx=−2(x2−1)(2x)−2(x2y−x−1)(2xy−1)
fy=−2x2(x2y−x−1)
{fx=0fy=0 If x=0, then
−2(0−1)(0)−2(0−0−1)(0−1)=−2=0 If x2y−x−1=0, then
−2(x2−1)(2x)−2(0)(2xy−1)=0 Since x=0, we take x1=−1,x2=1.
x=−1
(−1)2y−(−1)−1=0=>y=0 Point (−1,0).
x=1
(1)2y−1−1=0=>y=2 Point (1,2).
fxx=−12x2+2−12x2y2+12xy+4y
fxy=fyx=−8x3y+6x2+4x
fyy=−2x4 Point (−1,0)
fxx(−1,0)=−10<0
D=∣∣−1022−2∣∣=16>0There is a relative maximum at (−1,0).
Point (1,2)
fxx(1,2)=−26<0D=∣∣−26−6−6−2∣∣=16>0There is a relative maximum at (1,2).
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