Absolute extrema on a closed region
Problem 10.303 · medium
Find the absolute maximum and minimum values of \( \displaystyle f(x, y) = x^{2} - x y + 3 x - y^{2} + 3 y \) on the rectangle \( \displaystyle 0 \le x \le 2 \), \( \displaystyle 0 \le y \le 3 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - x y + 3 x - y^{2} + 3 y\right)\\\frac{\partial}{\partial y} \left(x^{2} - x y + 3 x - y^{2} + 3 y\right)\end{matrix}\right] = \left[\begin{matrix}2 x - y + 3\\- x - 2 y + 3\end{matrix}\right] \]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.
- \[ \left[\begin{matrix}0\\\frac{9}{4}\\0\\10\\\frac{41}{4}\\4\end{matrix}\right] \]f at every candidate: (0, 0), (0, 3/2), (0, 3), (2, 0), (2, 1/2), (2, 3).✓ Proved
- The largest value is 41/4, the smallest 0.
Answer \( \max = \frac{41}{4}\ \text{at}\ (2, \frac{1}{2});\ \min = 0\ \text{at}\ (0, 0), (0, 3) \)
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 (misleading) — The solution lists candidate points on the boundary but never justifies that there are no interior critical points (the gradient equations have no solution in the rectangle). A student might incorrectly assume that interior points were checked or that the listed points are exhaustive without this justification.qwen3.6:27b-mlx: fail (error) — The solution fails to find the interior critical point (x=0, y=3/2) by solving ∇f=0, incorrectly listing (0, 3/2) as a boundary point in the candidate list. While the final numerical answer is correct by coincidence, the method described is flawed and misleading.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to find the interior critical point (x=0, y=3/2) by solving ∇f=0, incorrectly listing (0, 3/2) as a boundary point in the candidate list. While the final numerical answer is correct by coincidence, the method described is flawed and misleading.gpt-oss:20b: fail (misleading) 2026-10-04 — The solution lists candidate points on the boundary but never justifies that there are no interior critical points (the gradient equations have no solution in the rectangle). A student might incorrectly assume that interior points were checked or that the listed points are exhaustive without this justification.qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to find the interior critical point (x=0, y=3/2), which yields a local maximum of 9/4. Although this value is not the global maximum, omitting it means the candidate list is incomplete and the method is incorrectly applied. Additionally, the global minimum is not 0; f(2,3) = -4, which is lower than 0.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.