Why is (0,0) a saddle point for

Question

x^4 + 2y^2 -4xy?

Answer 1

Maxima (Local maximum) and Minima (local minimum) and saddle point :

Theorem :

Let f be a function with two variables with continuous second order partial derivatives f_xx, f_yy and f_xy at critical point (a,b). Let

D = f_xx(a,b) f_yy(a,b) - f_xy²(a,b)

1) If D > 0 and f_xx(a,b) > 0, then f has a relative minimum at (a,b).

2) If D > 0 and f_xx(a,b) < 0, then f has a relative maximum at (a,b).

3) If D < 0, then f has a saddle point at (a,b).

4) If D = 0, then no conclusion can be drawn.

Step 1 :

The function F(x, y) = x⁴ + 2y² - 4xy.

First order partial derivatives.

f_x(x, y) = 4x³ - 4y

f_y(x, y) = 4y - 4x

Second order partial derivatives.

F_xx(x, y) = 12x²

F_yy(x, y) = 4

F_xy(x, y) = -4

Step 2 :

The critical points satisfy the equations f_x(x,y) = 0 and f_y(x,y) = 0.

So solve the following equations  F_x = 0  and F_y = 0 simultaneously.Hence

4x³ - 4y = 0

x³ - y = 0

y = x³

Substitute y = x³ in the equation f_y(x, y) = 4y - 4x = 0 .

4(x³) - 4x = 0

4x(x² - 1) = 0

4x = 0 and (x² - 1) = 0

x = 0 and x = ± 1.

Substitute x = 0  in equation y = x³ then y = 0.

Substitute x = 1 in equation y = x³  then y = 1.

Substitute x = -1 in equation y = x³  then y = -1.

Critical points are (0, 0), (1, 1) and (-1, -1).

Step 3 :

D = f_xx(x,y) f_yy(x,y) - f_xy²(x,y)

D = (12x² )(4) - (-4)²

D = 48x² - 16.

At (0,0)

D = 48(0)² - 16 = - 16 < 0

D < 0 , So (0, 0)  is saddle point.

Saddle point at (0,0).