Calculus

To find the stationary points, compute the first partial derivatives and equate to zero: $f_x = 3x^2 - 3y = 0 \implies y = x^2$ $f_y = 3y^2 - 3x = 0 \implies 3(x^2)^2 - 3x = 0 \implies 3x(x^3 - 1) = 0$ Thus, $x = 0$ or $x = 1$. - For $x = 0 \implies y = 0$, giving point $(0, 0)$. - For $x = 1 \implies y = 1$, giving point $(1, 1)$. Second partial derivatives: $r = f_{xx} = 6x, \quad s = f_{xy} = -3, \quad t = f_{yy} = 6y$ Discriminant $D = rt - s^2 = (6x)(6y) - (-3)^2 = 36xy - 9$. - At $(0,0)$: $D = 0 - 9 = -9 < 0$ (Saddle point). - At $(1,1)$: $D = 36(1)(1) - 9 = 27 > 0$ and $r = 6 > 0$ (Local minimum). The local minimum value is: $f(1, 1) = 1^3 + 1^3 - 3(1)(1) + 4 = 1 + 1 - 3 + 4 = 3$