Consider the function $f(x, y) = x^3 + y^3 - 3xy$. At the non-zero critical point $(x_0, y_0) = (1, 1)$, what is the local minimum value attained by $f(x,y)$?

Answer Key:

-1.0

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: First Partial Derivatives:** * $f_x = 3x^2 - 3y = 0 \implies y = x^2$. * $f_y = 3y^2 - 3x = 0 \implies x = y^2$. * Substituting gives $x = x^4 \implies x(x^3 - 1) = 0 \implies x = 0$ or $x = 1$. * Non-zero critical point is $(1, 1)$. 2. **Stage 2: Second Derivative (Hessian) Test:** * $f_{xx} = 6x$, $f_{yy} = 6y$, $f_{xy} = -3$. * At $(1,1)$: $f_{xx} = 6$, $f_{yy} = 6$, $f_{xy} = -3$. * Discriminant $D = f_{xx} f_{yy} - (f_{xy})^2 = (6)(6) - (-3)^2 = 36 - 9 = 27 > 0$. * Since $D > 0$ and $f_{xx} = 6 > 0$, the point $(1,1)$ is a true local minimum. 3. **Stage 3: Function Value at Minimum:** * $f(1, 1) = 1^3 + 1^3 - 3(1)(1) = 1 + 1 - 3 = -1.0$. 4. **Stage 4: Common Trap:** Forgetting the minus sign or entering the location (1) instead of the minimum value (-1). 5. **Stage 5: Official Reference:** James Stewart, *Calculus: Early Transcendentals*, Chapter 14.

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