Locate all relative extreme and saddle points of f(x,y)=4xy-x^4-y^4.?

Question

Locate all relative extreme and saddle points

Answer 1

∂f/∂x=4y-4x^3; ∂f/∂y=4x-4y^3. ∂f/∂x=0 and ∂f/∂y=0 at turning points, so 4y=4x^3 and 4x=4y^3, y=x^3 and x=y^3.

x=x^9 by substitution so x(1-x^8)=0. The (real) solutions are 0, 1, -1 leading to 0, 1 -1 for y.

x=y=0, x=y=1,  x=y=-1 satisfy these conditions simultaneously. So f(x,y)=0=f(0,0), f(x,y)=2=f(1,1) and f(x,y)=2=f(-1,-1).

The points are (x,y,f)=(0,0,0), (1,1,2), (-1,-1,2).

Answer 2

f(x,y) = 4xy - x^4 - y^4

        = <4y - 4x^3,4x - 4y^3>

We solve:

4y - 4x^3 = 0 so that y = x^3

Hence 4x - 4(x^3)^3 = 0

x - x^9 = 0 so that x = 1 or x = 0 or x = -1

this gives us the points (1,1), (0,0) and (-1,-1)

We have fxx = -12x^2, fyy = -12y^2, and fxy = 4

Hence D = 144x^2y^2 - 16
max = (1,1)
Saddle => (0,0)
max =>(-1,-1)

Finding Maxima & Minima