Given that x+2y=4,determine the minimum value of x^2+xy-y^2.
x = 4-2y
V = value =(4-2y)^2 + (4-2y)y - y^2
V = 16 - 16 y + 4y^2 + 4 y -2 y^2 -y^2
V = y^2 - 12 y + 16
so
y^2 - 12 y = V-16
y^2 - 12 y + 6^2 = V - 16 + 36
(Y-6)^2 = V+20
vertex at v = -20