Question #334437

f(x,y)= -(x2-1)2-(x2y-x-1)2

Find the critical points of f (x, y) and Using the Second Partials Test (SPT), find the relative extrema’s of f (x ,y)


1
Expert's answer
2022-04-28T10:25:44-0400
f(x,y)=(x21)2(x2yx1)2f(x,y)= -(x^2-1)^2-(x^2y-x-1)^2

Find the critical point(s)


fx=2(x21)(2x)2(x2yx1)(2xy1)f_x=-2(x^2-1)(2x)-2(x^2y-x-1)(2xy-1)

fy=2x2(x2yx1)f_y=-2x^2(x^2y-x-1)

{fx=0fy=0\begin{cases} f_x=0 \\ f_y=0 \end{cases}

If x=0,x=0, then


2(01)(0)2(001)(01)=20-2(0-1)(0)-2(0-0-1)(0-1)=-2\not=0

If x2yx1=0,x^2y-x-1=0, then


2(x21)(2x)2(0)(2xy1)=0-2(x^2-1)(2x)-2(0)(2xy-1)=0

Since x0,x\not=0, we take x1=1,x2=1.x_1=-1, x_2=1.

x=1x=-1


(1)2y(1)1=0=>y=0(-1)^2y-(-1)-1=0=>y=0

Point (1,0).(-1, 0).


x=1x=1

(1)2y11=0=>y=2(1)^2y-1-1=0=>y=2

Point (1,2).(1, 2).

fxx=12x2+212x2y2+12xy+4yf_{xx}=-12x^2+2-12x^2y^2+12xy+4y

fxy=fyx=8x3y+6x2+4xf_{xy}=f_{yx}=-8x^3y+6x^2+4x

fyy=2x4f_{yy}=-2x^4

Point (1,0)(-1, 0)

fxx(1,0)=10<0f_{xx}(-1,0)=-10<0


D=10222=16>0D=\begin{vmatrix} -10 & 2 \\ 2 & -2 \end{vmatrix}=16>0

There is a relative maximum at (1,0).(-1, 0).


Point (1,2)(1, 2)

fxx(1,2)=26<0f_{xx}(1,2)=-26<0D=26662=16>0D=\begin{vmatrix} -26 & -6 \\ -6 & -2 \end{vmatrix}=16>0

There is a relative maximum at (1,2).(1, 2).



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