f(x,y)=x^3-2xy+y^2+5 critical points

Question

Determine the critical points. relative max., relative min., or saddle point using second partials test

Answer 1

f(x,y)=x³-2xy+y²+5; f_x=3x²-2y; f_y=-2x+2y; f_xx=6x; f_yy=2; f_xy=-2=f_yx.

At critical points f_x=0 and f_y=0 at the same point:

3x²-2y=0, -2x+2y=0, so y=x and 3x²-2x=0, x(3x-2)=0⇒x=y=0, x=y=⅔.

f(0,0)=5; f(⅔,⅔)=131/27.

Critical points are (0,0,5) and (⅔,⅔,131/27).

For second derivative test D(x,y)=f_xx.f_yy-(f_xy)²:

D(0,0)=-4<0 (saddle-point), D(⅔,⅔)=8-4>0 and f_xx>0 (minimum).

(0,0,5) is a saddle-point; (⅔,⅔,131/27) is a minimum.