Home›Calculus 3›Absolute extrema on a closed region›Problem 10.560

Absolute extrema on a closed region

Problem 10.560 · medium

Find the absolute maximum and minimum values of \( \displaystyle f(x, y) = x^{2} + x y - 3 x - 2 y^{2} + y \) on the rectangle \( \displaystyle -2 \le x \le 1 \), \( \displaystyle 0 \le y \le 3 \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + x y - 3 x - 2 y^{2} + y\right)\\\frac{\partial}{\partial y} \left(x^{2} + x y - 3 x - 2 y^{2} + y\right)\end{matrix}\right] = \left[\begin{matrix}2 x + y - 3\\x - 4 y + 1\end{matrix}\right] \]
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.
\[ \left[\begin{matrix}10\\-11\\-15\\-2\\- \frac{3}{2}\\-14\end{matrix}\right] \]
f at every candidate: (-2, 0), (-2, 3), (0, 3), (1, 0), (1, 1/2), (1, 3).✓ Proved
The largest value is 10, the smallest -15.

Answer \( \max = 10\ \text{at}\ (-2, 0);\ \min = -15\ \text{at}\ (0, 3) \)

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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to identify the interior critical point (2, 1/2), which is the only solution to ∇f = 0. Although this point lies outside the domain and thus does not affect the final numerical answer, omitting it constitutes an incomplete application of the Extreme Value Theorem procedure for finding absolute extrema.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-13
qwen3.6:27b-mlx: fail (style) 2026-10-13 — [domain objection, downgraded to style] The solution fails to identify the interior critical point (2, 1/2), which is the only solution to ∇f = 0. Although this point lies outside the domain and thus does not affect the final numerical answer, omitting it constitutes an incomplete application of the Extreme Value Theorem procedure for finding absolute extrema.
gpt-oss:20b: pass 2026-10-13
qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution fails to find the interior critical point, which is the global minimum. Solving ∇f = 0 yields (x, y) = (1, 1/2), where f(1, 1/2) = -3.5. Although this point is listed in the candidate values in line 3, the text in line 2 implies checking only edges and corners, omitting the interior. Furthermore, the stated answer claims the minimum is -15, but f(1, 1/2) = -3.5 is not the minimum; however, checking the corners: f(-2,0)=10, f(-2,3)=-11, f(0,3)=-15, f(1,0)=-2. Wait, f(1, 1/2) = 1 + 0.5 - 3 - 0.5 + 0.5 = -1.5? No. f(1, 0.5) = 1^2 + 1(0.5) - 3(1) - 2(0.5)^2 + 0.5 = 1 + 0.5 - 3 - 0.5 + 0.5 = -1.5. Let's re-calculate. f(1, 1/2) = 1 + 0.5 - 3 - 0.5 + 0.5 = -1.5. The value -3/2 in line 3 corresponds to (1, 1/2). Is -15 the minimum? f(0,3) = 0 + 0 - 0 - 18 + 3 = -15. f(-2, 3) = 4 - 6 + 6 - 18 + 3 = -11. f(1, 3) = 1 + 3 - 3 - 18 + 3 = -14. f(1, 0) = 1 + 0 - 3 - 0 + 0 = -2. f(-2, 0) = 4 + 0 + 6 - 0 + 0 = 10. So min is -15. The interior point is a saddle or local max? Hessian: f_xx=2, f_yy=-4, f_xy=1. D = 2(-4) - 1^2 = -7 < 0, so saddle point. So the interior critical point is not a min or max. The solution lists it but doesn't explicitly state it's a saddle or why it's not the extremum, but more importantly, line 2 says 'On each edge... include corners', completely omitting the instruction to check interior critical points, which is a standard part of the method. While the final answer is correct because the interior point is a saddle, the method described in sentence 2 is incomplete/wrong for finding absolute extrema on a closed region, as it ignores the interior. A student following this advice would miss interior 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 2026-10-13 with SymPy 1.14.0.