Classify all critical points for f (x,y) = 2x^2 - y^3 - 2xy.

Answer 1

f'_x = 4x - 2y = 0;

f'_y = - 3y² - 2x = 0

From 1st equation y = 2x. So, - 12x² - 2x = 0, x = 0 and x = - 1/6, y = 0 and y = - 1/3.

We have 2 critical points: (0, 0) and (- 1/6, - 1/3).

To classify critical points let calculate secon partial derivatives: f'''_xx = 4, f''_xy= - 2 and f''_yy= - 6y.

For critical point (0,0): f''_xx·f''_yy - f''_xy² = 4(-6·0) - (- 2)² = - 4 < 0. So at (0,0) We have max; max f = f(0,0) = 0

For critical point (-1/6, - 1/3): f''_xx·f''_yy - f''_xy² = 4(- 6·(- 1/3)) - (- 2)² = 8 - 4 = 4 > 0. So at this point we have min:

min f = f(- 1/6, - 1/3) = 2(- 1/6)² - (- 1/3)³ - 2(- 1/6)(- 1/3) = 1/18 + 1/27 - 1/9 = (3 + 2 - 6)/54 = - 1/54