find critical point $f(x,y) = 8 + 2y^3+2x^3 - 3xy$

Question

$f(x,y) = 8 + 2y^3+2x^3 - 3xy$

$f_x = 6x^2 - 3y$

$f_{xx} = 12x$, $x= 0$, $y=0$

$f_y = 6y^2 -3x$

$f_{yy}=12y$

$f_{xy}=-3$

at $(0,0)$, $D = -9 < 0$

Is it one saddle point $(0,0)$?

Answer 1

You need to first find ALL the points where $f_x = f_y = 0$.

$f_x = 6x^2 - 3y = 0$ gives us $y = 2x^2$ (1)

$f_y = 6y^2 - 3x = 0$ gives us $x = 2y^2$ (2)

Substitute (1) into (2) to get $x = 2(2x^2) = 8x^4$, i.e. $8x^4-x = 0$.

This has another real root besides $x = 0$, which will give you another critical point.