By the function f(x,y)=4+x^3+y^3−3xy.

Question

State whether each is a local minimum, local maximum or saddle point.?

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) = x³ + y³ - 3xy +4

First order partial derivatives.

f_x(x, y) = 3x² - 3y

f_y(x, y) = 3y² - 3x

Second order partial derivatives.

F_xx(x, y) = 6x

F_yy(x, y) = 6y

F_xy(x, y) = -3

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

3x² - 3y = 0

3x² = 3y

y = x²

3y² - 3x = 0

Substitute y = x²

3(x²)² - 3x = 0

x⁴ - x = 0

x( x³ - 1 ) = 0

x = 0  and   x³ - 1 = 0

x = 0  and   x = 1

Substitute x = 0  in equation y = x² then y = 0

Substitute x = 1 in equation y = x²  then y = 1

Critical points are (0,0) , (1,1)

Step 3 :

D = f_xx(x,y) f_yy(x,y) - f_xy²(x,y)

D = (6x)(6y) - (-3)²

D = 36xy - 9

At (0,0)

D = 36(0)(0)-9 = -9 < 0

D < 0 , So ( 0 , 0 )  is saddle point

At (1,1)

D = 36(1)(1)-9 = 27 > 0

F_xx = 6x = 6(1) = 6 >0

D > 0 and F_xx > 0 So f has a relative minimum at (1,1).

Solution :

Relative minimum at (1,1)

Saddle point at (0,0).