Lagrange multipliers
Problem 10.163 · medium Video
Use Lagrange multipliers to find the maximum and minimum of \( \displaystyle f(x, y) = 4 x + 3 y \) on the circle \( \displaystyle x^2 + y^2 = 1 \).
- Solve ∇f = λ∇g with g(x, y) = x² + y² − r² = 0.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(4 x + 3 y\right)\\\frac{\partial}{\partial y} \left(4 x + 3 y\right)\end{matrix}\right] = \left[\begin{matrix}4\\3\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + y^{2} - 1\right)\\\frac{\partial}{\partial y} \left(x^{2} + y^{2} - 1\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\2 y\end{matrix}\right] \]∇g.✓ Proved
- So x = 4/(2λ), y = 3/(2λ); substituting into the constraint gives λ = ±√(25)/(2√1).
- \[ 5 \]The maximum; the minimum is its negative.
Answer \( \max = 5,\ \min = -5 \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to explicitly calculate the maximum and minimum values of f(x,y) by substituting the critical points back into f. It jumps from finding lambda to stating the answer '5' without showing the intermediate step of evaluating f at the constrained points, which is the actual goal of the problem.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly calculate the maximum and minimum values of f(x,y) by substituting the critical points back into f. It jumps from finding lambda to stating the answer '5' without showing the intermediate step of evaluating f at the constrained points, which is the actual goal of the problem.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly calculate the coordinates (x, y) corresponding to the Lagrange multipliers, which is necessary to rigorously determine which lambda value corresponds to the maximum and which to the minimum. While the final numerical answer is correct, the logical step 'the minimum is its negative' relies on symmetry intuition rather than the explicit evaluation of f at the critical points derived from lambda, which is the standard procedure for Lagrange multipliers.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly calculate the minimum value, stating only that it is the negative of the maximum without providing the value -5. Additionally, the intermediate step for lambda is presented with ambiguous notation and lacks the explicit substitution steps to derive the coordinates.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.