Absolute extrema on a closed region
Problem 10.358 · medium
Find the absolute maximum and minimum values of \( \displaystyle f(x, y) = 2 x^{2} + 4 x + y^{2} + y \) on the rectangle \( \displaystyle -2 \le x \le 1 \), \( \displaystyle -1 \le y \le 2 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(2 x^{2} + 4 x + y^{2} + y\right)\\\frac{\partial}{\partial y} \left(2 x^{2} + 4 x + y^{2} + y\right)\end{matrix}\right] = \left[\begin{matrix}4 x + 4\\2 y + 1\end{matrix}\right] \]Interior critical points solve ∇f = 0.✓ Proved
- On each edge f is a function of one variable: find its critical points there too, and include the four corners.
- \[ \left[\begin{matrix}0\\- \frac{1}{4}\\6\\-2\\- \frac{9}{4}\\4\\6\\\frac{23}{4}\\12\end{matrix}\right] \]f at every candidate: (-2, -1), (-2, -1/2), (-2, 2), (-1, -1), (-1, -1/2), (-1, 2), (1, -1), (1, -1/2), (1, 2).✓ Proved
- The largest value is 12, the smallest -9/4.
Answer \( \max = 12\ \text{at}\ (1, 2);\ \min = - \frac{9}{4}\ \text{at}\ (-1, - \frac{1}{2}) \)
Lines: 2 proved, 2 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f sampled on a 241 × 241 grid never beats the claimed max or min, and comes within 5% of both |
Reviewers
gpt-oss:20b: fail (error) — The interior critical point is at (-1,-½), but f(-1,-½)=−7/4, not −9/4. Moreover, the global minimum on the rectangle is f(−1,−1)=−2, not −9/4. The candidate list and evaluation are incorrect.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to explicitly identify the interior critical point (-1, -1/2) and verify that it lies within the domain before including it in the candidate list. While the point is present in the final list, the logical step of solving ∇f=0 and checking the domain is missing, making the derivation of the minimum value unjustified.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-10-05 — [domain objection, downgraded to style] The solution fails to explicitly identify the interior critical point (-1, -1/2) and verify that it lies within the domain before including it in the candidate list. While the point is present in the final list, the logical step of solving ∇f=0 and checking the domain is missing, making the derivation of the minimum value unjustified.gpt-oss:20b: fail (error) 2026-10-05 — The interior critical point is at (-1,-½), but f(-1,-½)=−7/4, not −9/4. Moreover, the global minimum on the rectangle is f(−1,−1)=−2, not −9/4. The candidate list and evaluation are incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to identify the interior critical point (-1, -1/2) as a candidate in step 1, although it is included in the list in step 3. This makes the logical flow disjointed and suggests the student does not understand that interior critical points are distinct from boundary candidates.gpt-oss:20b: pass 2026-10-05
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/absolute_extrema_2var, checked 2026-10-05 with SymPy 1.14.0.