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Classic optimization: wire, window, fence

Problem 3.678 · medium

A rectangular pen of area \( \displaystyle 600 \) m² is split in two by a fence parallel to two of its sides (\( \displaystyle x \) is the length of those sides). What \( \displaystyle x \) uses the least fencing, and how much is that?

With sides x and 600/x, the fence is three x-sides and two others: F(x) = 3x + 2·600/x.
Reviewed
\[ \frac{d}{d x} \left(3 x + \frac{1200}{x}\right) = 3 - \frac{1200}{x^{2}} \]
Differentiate.✓ Proved
\[ \left. 3 - \frac{1200}{x^{2}} \right|_{\substack{ x=20 }} = 0 \]
The only critical point in the domain is x = 20.✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \left(3 x + \frac{1200}{x}\right) \right|_{\substack{ x=20 }} = \frac{3}{10} \]
The second derivative is positive: a minimum.✓ Proved
\[ \left. 3 x + \frac{1200}{x} \right|_{\substack{ x=20 }} = 120 \]
The fencing there.✓ Proved

Answer \( x = 20,\ \text{fencing} = 120 \)

Lines: 4 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the objective sampled at 200,000 points across the domain

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly models the fencing problem, identifies the critical point, verifies it is a minimum using the second derivative test, and calculates the correct minimum fencing amount.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly models the fencing problem, identifies the critical point, verifies it is a minimum using the second derivative test, and calculates the correct minimum fencing amount.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly models the problem, identifies the critical point, verifies it is a minimum using the second derivative test, and calculates the correct minimum fencing amount.
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly models the fencing problem, identifies the critical point, verifies it is a minimum using the second derivative test, and calculates the correct final values.
gpt-oss:20b: fail (error) 2026-10-14 — The setup assumes the internal fence is parallel to the sides of length x, giving F(x)=3x+1200/x. In fact, if the fence is parallel to the sides of length x, the internal fence has length 600/x, so the correct total fencing is 2x+1800/x. The solution’s model is therefore incorrect.

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/optimization_classic, checked 2026-10-17 with SymPy 1.14.0.