Local extrema and saddle points
Problem 10.252 · easy Video
Find and classify the critical points of \( \displaystyle f(x, y) = - x^{2} + 6 x - 2 y^{2} - 12 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} + 6 x - 2 y^{2} - 12\right)\\\frac{\partial}{\partial y} \left(- x^{2} + 6 x - 2 y^{2} - 12\right)\end{matrix}\right] = \left[\begin{matrix}6 - 2 x\\- 4 y\end{matrix}\right] \]The partial derivatives.✓ Proved
- \[ \left[\begin{matrix}0\\0\end{matrix}\right] \]Both vanish at (3, 0), the only solution.
- \[ 8 \]D = f_xx f_yy − f_xy².
- D = 8 > 0 with f_xx < 0: a local maximum.
Answer \( \text{local maximum at } (3, 0) \)
Lines: 1 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 3 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 4 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f on a small circle around the point is compared with its centre value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The discriminant D is calculated as 8, but for f(x,y) = -x^2 + 6x - 2y^2 - 12, f_xx = -2 and f_yy = -4, so D = (-2)(-4) - 0 = 8. However, the solution claims D = 8 > 0 with f_xx < 0 implies a local maximum. While the conclusion is correct, the value of D is actually 8, not just '8' in a vacuum, but more importantly, the standard second derivative test requires D > 0 and f_xx < 0 for a local maximum. The calculation D=8 is correct. Wait, let me re-read carefully. f_xx = -2, f_yy = -4, f_xy = 0. D = (-2)(-4) - 0 = 8. D > 0 and f_xx < 0 implies local maximum. The logic is sound. Why did I think it was wrong? Ah, I see no error. Let me double check the problem statement. f(x,y) = -x^2 + 6x - 2y^2 - 12. Critical point: fx = -2x + 6 = 0 => x=3. fy = -4y = 0 => y=0. Point (3,0). fxx = -2, fyy = -4, fxy = 0. D = 8. D > 0, fxx < 0 => Local Max. The solution is correct. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The discriminant D is calculated as 8, but for f(x,y) = -x^2 + 6x - 2y^2 - 12, f_xx = -2 and f_yy = -4, so D = (-2)(-4) - 0 = 8. However, the solution claims D = 8 > 0 with f_xx < 0 implies a local maximum. While the conclusion is correct, the value of D is actually 8, not just '8' in a vacuum, but more importantly, the standard second derivative test requires D > 0 and f_xx < 0 for a local maximum. The calculation D=8 is correct. Wait, let me re-read carefully. f_xx = -2, f_yy = -4, f_xy = 0. D = (-2)(-4) - 0 = 8. D > 0 and f_xx < 0 implies local maximum. The logic is sound. Why did I think it was wrong? Ah, I see no error. Let me double check the problem statement. f(x,y) = -x^2 + 6x - 2y^2 - 12. Critical point: fx = -2x + 6 = 0 => x=3. fy = -4y = 0 => y=0. Point (3,0). fxx = -2, fyy = -4, fxy = 0. D = 8. D > 0, fxx < 0 => Local Max. The solution is correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-09-29gpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: pass 2026-09-29gpt-oss:20b: pass 2026-09-29
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/critical_points_2var, checked 2026-10-13 with SymPy 1.14.0.