f(x,y)=3x²-3xy+2y²-9x-3y-11.
∂f/∂x=6x-3y-9=0 at an extremum (1)
∂f/∂y=-3x+4y-3=0 at an extremum (2)
∂²f/∂x²=6, ∂²f/∂y²=4, ∂f/∂x∂y=-3
Solve these simultaneous equations:
(1)+2(2): 5y-15=0, 5y=15, y=3.
So 6x-9-9=0, 6x=18, x=3.
Now we need to test what type of extremum (3,3) is.
D=(∂²f/∂x²)(∂²f/∂y²)-(∂f/∂x∂y)² at (3,3):
D=24-9=15.
Since D and ∂²f/∂x² are both >0, there is a local minimum at (3,3).
f(3,3)=27-27+18-27-9-11=-29.
There is a local minimum at (3,3,-29).