Suppose vertices are O(0,0),A(2,0),B(2,2)
For interior points with respect to x:
f(x)=(3x^2)-3y (a)
Interior points with respect y:
f(y)3-3x (b)
for the boundary points we take on the segment OA
y=0
U(x)=f(x,0)=x^3
Regarded as function of x denied on the closed interval (2<=x<=2) its extreme value may be occur at the end point x=2 and so on which corresponded to the point (2,0) and (2,2)
U’(x)=3x^2=0
On segment OB x=0 and V(y)=(0,y)=3y
Using symmetric form possible points are also these (0,0)(2,0)(2,2)
The interior point of AB y=2-x
Fy=x^3+3(2-x)-3x(2-x)
Fy=(x^3)-3(x^2)-9x+6
Take derivative
Fy’=3x^2-7x
Points are (2,2)