Lagrange multipliers
Problem 10.122 · medium Video
Use Lagrange multipliers to find the maximum and minimum of \( \displaystyle f(x, y) = 3 x + 2 y \) on the circle \( \displaystyle x^2 + y^2 = 4 \).
- Solve ∇f = λ∇g with g(x, y) = x² + y² − r² = 0.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(3 x + 2 y\right)\\\frac{\partial}{\partial y} \left(3 x + 2 y\right)\end{matrix}\right] = \left[\begin{matrix}3\\2\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + y^{2} - 4\right)\\\frac{\partial}{\partial y} \left(x^{2} + y^{2} - 4\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\2 y\end{matrix}\right] \]∇g.✓ Proved
- So x = 3/(2λ), y = 2/(2λ); substituting into the constraint gives λ = ±√(13)/(2√4).
- \[ 2 \sqrt{13} \]The maximum; the minimum is its negative.
Answer \( \max = 2 \sqrt{13},\ \min = - 2 \sqrt{13} \)
Lines: 2 proved, 3 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 | Not checked | — | a sentence; read, not computed |
| 5 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f around 100,000 points of the circle tops out at the same value |
Reviewers
gpt-oss:20b: fail (error) — The solution incorrectly computes the extreme values of f. Substituting x=3/(2λ), y=1/λ into the constraint yields λ=±√13/4, but then f=3x+2y=(13/2)/λ, giving f=±4√13, not ±2√13. The conclusion is mathematically incorrect.qwen3.6:27b-mlx: pass — The solution correctly sets up the Lagrange multiplier equations, solves for the critical points, and identifies the correct maximum and minimum values.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly sets up the Lagrange multiplier equations, solves for the critical points, and identifies the correct maximum and minimum values.gpt-oss:20b: fail (error) 2026-09-28 — The solution incorrectly computes the extreme values of f. Substituting x=3/(2λ), y=1/λ into the constraint yields λ=±√13/4, but then f=3x+2y=(13/2)/λ, giving f=±4√13, not ±2√13. The conclusion is mathematically incorrect.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the method of Lagrange multipliers, derives the correct critical points, and identifies the maximum and minimum values accurately.gpt-oss:20b: fail (error) 2026-09-27 — The algebra in the substitution step is incorrect: λ is found to be ±√5/(2√2), not ±√13/(2√4). Consequently the computed extrema are wrong; the correct maximum and minimum are ±13√(2/5), not ±2√13.qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the method of Lagrange multipliers, derives the correct critical values for lambda, and identifies the correct maximum and minimum values.gpt-oss:20b: pass 2026-09-26
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/lagrange_multipliers, checked 2026-10-13 with SymPy 1.14.0.