Home›Calculus 3›Absolute extrema on a closed region›Problem 10.455

Absolute extrema on a closed region

Problem 10.455 · 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 \( \displaystyle -1 \le x \le 2 \), \( \displaystyle -1 \le y \le 2 \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} - x y - 3 x + y^{2} + 3 y\right)\\\frac{\partial}{\partial y} \left(- x^{2} - x y - 3 x + y^{2} + 3 y\right)\end{matrix}\right] = \left[\begin{matrix}- 2 x - y - 3\\- x + 2 y + 3\end{matrix}\right] \]
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.
\[ \left[\begin{matrix}-1\\14\\-10\\- \frac{41}{4}\\-4\end{matrix}\right] \]
f at every candidate: (-1, -1), (-1, 2), (2, -1), (2, -1/2), (2, 2).✓ Proved
The largest value is 14, the smallest -41/4.

Answer \( \max = 14\ \text{at}\ (-1, 2);\ \min = - \frac{41}{4}\ \text{at}\ (2, - \frac{1}{2}) \)

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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to find the interior critical point at (-2, 0), which lies outside the domain, but more critically, it omits checking the critical points on the edges x=-1 and y=-1. For example, on the edge x=-1, f(-1, y) = y^2 + 4y - 2, which has a critical point at y=-2 (outside) but the endpoints are checked. However, on the edge y=2, f(x, 2) = -x^2 - 5x + 2, critical point at x=-2.5 (outside). On edge x=2, f(2, y) = y^2 + y - 10, critical point at y=-1/2 (inside, checked). On edge y=-1, f(x, -1) = -x^2 - 2x - 4, critical point at x=-1 (corner, checked). Wait, let's re-evaluate. The interior critical point is (-2, 0), which is outside. The edge critical points: x=-1 -> y=-2 (out); x=2 -> y=-1/2 (in, checked); y=-1 -> x=-1 (corner, checked); y=2 -> x=-2.5 (out). So the candidates listed are actually sufficient. However, the sentence in step 2 says 'find its critical points there too', but step 3 only lists 5 points. Did it miss any? Let's check y=-1 edge again. f(x, -1) = -x^2 - x(-1) - 3x + (-1)^2 + 3(-1) = -x^2 + x - 3x + 1 - 3 = -x^2 - 2x - 2. Derivative -2x - 2 = 0 => x = -1. This is a corner. So no new interior edge points. The solution seems correct in its candidates. Why did I think it failed? Let me re-read the problem statement and answer. Max 14 at (-1, 2). Min -41/4 at (2, -1/2). Let's check f(2, -1/2) = -4 - 2(-1/2) - 6 + 1/4 - 3/2 = -4 + 1 - 6 + 0.25 - 1.5 = -10.25 = -41/4. Correct. Let's check f(-1, 2) = -1 - (-2) - (-3) + 4 + 6 = -1 + 2 + 3 + 4 + 6 = 14. Correct. Are there any other points? The interior critical point is (-2, 0), outside. Edge critical points: x=-1, y=-2 (out); x=2, y=-1/2 (in); y=-1, x=-1 (corner); y=2, x=-2.5 (out). So the set of candidates is complete. The solution is actually correct. Why would it fail? Maybe the sentence 'On each edge f is a function of one variable...' is too brief? No, it's a standard method. Is there a subtle error? The problem asks for absolute max/min. The solution finds them. I will mark it pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: fail (style) 2026-10-09 — [domain objection, downgraded to style] The solution fails to find the interior critical point at (-2, 0), which lies outside the domain, but more critically, it omits checking the critical points on the edges x=-1 and y=-1. For example, on the edge x=-1, f(-1, y) = y^2 + 4y - 2, which has a critical point at y=-2 (outside) but the endpoints are checked. However, on the edge y=2, f(x, 2) = -x^2 - 5x + 2, critical point at x=-2.5 (outside). On edge x=2, f(2, y) = y^2 + y - 10, critical point at y=-1/2 (inside, checked). On edge y=-1, f(x, -1) = -x^2 - 2x - 4, critical point at x=-1 (corner, checked). Wait, let's re-evaluate. The interior critical point is (-2, 0), which is outside. The edge critical points: x=-1 -> y=-2 (out); x=2 -> y=-1/2 (in, checked); y=-1 -> x=-1 (corner, checked); y=2 -> x=-2.5 (out). So the candidates listed are actually sufficient. However, the sentence in step 2 says 'find its critical points there too', but step 3 only lists 5 points. Did it miss any? Let's check y=-1 edge again. f(x, -1) = -x^2 - x(-1) - 3x + (-1)^2 + 3(-1) = -x^2 + x - 3x + 1 - 3 = -x^2 - 2x - 2. Derivative -2x - 2 = 0 => x = -1. This is a corner. So no new interior edge points. The solution seems correct in its candidates. Why did I think it failed? Let me re-read the problem statement and answer. Max 14 at (-1, 2). Min -41/4 at (2, -1/2). Let's check f(2, -1/2) = -4 - 2(-1/2) - 6 + 1/4 - 3/2 = -4 + 1 - 6 + 0.25 - 1.5 = -10.25 = -41/4. Correct. Let's check f(-1, 2) = -1 - (-2) - (-3) + 4 + 6 = -1 + 2 + 3 + 4 + 6 = 14. Correct. Are there any other points? The interior critical point is (-2, 0), outside. Edge critical points: x=-1, y=-2 (out); x=2, y=-1/2 (in); y=-1, x=-1 (corner); y=2, x=-2.5 (out). So the set of candidates is complete. The solution is actually correct. Why would it fail? Maybe the sentence 'On each edge f is a function of one variable...' is too brief? No, it's a standard method. Is there a subtle error? The problem asks for absolute max/min. The solution finds them. I will mark it pass.
gpt-oss:20b: fail (error) 2026-10-09 — The solution omits the interior critical point (x,y)=(-3/5,-9/5) which lies outside the rectangle, but it also fails to check the edge x=-1, y=-1/2 and the edge y=-1, x=2/5, where the true extrema occur. Consequently the candidate list is incomplete and the stated maximum/minimum are incorrect.
qwen3.6:27b-mlx: fail (style) 2026-10-09 — [domain objection, downgraded to style] The solution fails to evaluate the function at the interior critical point (x, y) = (-1/2, -1/2), which lies within the domain. Although the global extrema happen to occur on the boundary, omitting the interior critical point from the candidate list is a procedural error in finding absolute extrema.

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 2026-10-09 with SymPy 1.14.0.