Find the min and max VALUES of f(x,y) = x^2y on g(x,y) = x^2+y^2 = 3

Answer:
F(x,y,lambda) = x^2y - lambda(x^2+y^2-3)
F_x = 2xy - 2 lambda x
F_y = x^2 - 2 lambda y
F_lambda = -(x^2+y^2-3) 

From F_x = 0 we get x = 0 or y = lambda
From x = 0, we get y = +/- sqrt(3) and lambda = 0
From y = lambda we get x^2 = 2 y^2 so 3y^2 = 3, y = +/- 1 and x = +/- sqrt(2)

f(0,  sqrt(3))  =  0
f(0, -sqrt(3))  =  0
f(sqrt(2),  1)  =  2 <-- MAX VALUE
f(sqrt(2), -1)  = -2 <-- MIN VALUE
f(-sqrt(2), 1)  =  2
f(-sqrt(2),-1)  = -2