Finding the Critical Points

For the function

f(x,y) = 3xy - x³ - y³

we can calculate

$$\mbox{grad} f(x,y) = 3(y - x^2)\hat{i} + 3(x - y^2) \hat{j}.$$ (5)

Thus, critical points will occur at the solutions to the system of equations

$$3(y - x^2) = 0 \qquad \qquad 3(x - y^2) = 0.$$ (6)

Using the first of these equations to eliminate y we find that

$$3(x - x^4) = 0 \Rightarrow x (x- 1)(x^2 + x + 1) = 0.$$ (7)

The only real solutions to this equation are x = 0 and x = 1. Using this, we find that, since y = x², the two critical points are (0,0) and (1,1).