Classic optimization: wire, window, fence
Problem 3.666 · medium
A rectangular pen of area \( \displaystyle 2400 \) 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 2400/x, the fence is three x-sides and two others: F(x) = 3x + 2·2400/x.
- \[ \frac{d}{d x} \left(3 x + \frac{4800}{x}\right) = 3 - \frac{4800}{x^{2}} \]Differentiate.✓ Proved
- \[ \left. 3 - \frac{4800}{x^{2}} \right|_{\substack{ x=40 }} = 0 \]The only critical point in the domain is x = 40.✓ Proved
- \[ \left. \frac{d^{2}}{d x^{2}} \left(3 x + \frac{4800}{x}\right) \right|_{\substack{ x=40 }} = \frac{3}{20} \]The second derivative is positive: a minimum.✓ Proved
- \[ \left. 3 x + \frac{4800}{x} \right|_{\substack{ x=40 }} = 240 \]The fencing there.✓ Proved
Answer \( x = 40,\ \text{fencing} = 240 \)
Lines: 4 proved, 1 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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: fail (error) — The derivative zero condition gives x^2=4800, so the critical point is x≈69.3, not 40. The solution’s algebraic step 3 is incorrect, leading to an incorrect minimum fencing value.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: fail (error) 2026-10-17 — The derivative zero condition gives x^2=4800, so the critical point is x≈69.3, not 40. The solution’s algebraic step 3 is incorrect, leading to an incorrect minimum fencing value.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 minimum fencing amount.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly models the fencing cost function based on the problem statement and accurately applies calculus to find the minimum. The steps are logically sound and the final answer is correct.gpt-oss:20b: pass 2026-10-14
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.