Find the nature of the stationary point of the surface $f(x,y) = x^3 + y^3 - 3xy$ at $(x,y) = (1, 1)$:

(A) Local Minimum

(B) Local Maximum

(C) Saddle Point

(D) Inconclusive

Answer Key:

A

### Solution: $f_x = 3x^2 - 3y$, $f_y = 3y^2 - 3x$. Second derivatives: $f_{xx} = 6x$, $f_{yy} = 6y$, $f_{xy} = -3$. At $(1,1)$: $r = f_{xx} = 6$, $t = f_{yy} = 6$, $s = f_{xy} = -3$. Discriminant $rt - s^2 = (6)(6) - (-3)^2 = 36 - 9 = 27 > 0$. Since $r = 6 > 0$ and $rt - s^2 > 0$, $(1,1)$ is a local minimum. Correct: A.

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