Find the absolute maximum and minimum values of f on the set D. f(x,y)=x^3 -3x -y^3 +12y+6, D is quadrilateral whose vertices are (-2,3), (2,3), (2,2), (-2,-2).

Answer 1

∂f/∂x = 3x^2-3 = 3(x^2-1)
∂f/∂y = -3y^2 + 12 = 3(4-y^2)
So any relative extrema are at (±1,±2) which are within D.
Now you just have to find extrema along the boundaries:
y=3
x=±2
y=x