Lagrange multipliers
Problem 10.230 · medium Video
Use Lagrange multipliers to find the maximum and minimum of \( \displaystyle f(x, y) = 3 x + 4 y \) on the circle \( \displaystyle x^2 + y^2 = 4 \).
- Solve ∇f = λ∇g with g(x, y) = x² + y² − r² = 0.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(3 x + 4 y\right)\\\frac{\partial}{\partial y} \left(3 x + 4 y\right)\end{matrix}\right] = \left[\begin{matrix}3\\4\end{matrix}\right] \]∇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 = 4/(2λ); substituting into the constraint gives λ = ±√(25)/(2√4).
- \[ 10 \]The maximum; the minimum is its negative.
Answer \( \max = 10,\ \min = -10 \)
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 (misleading) — The final sentence asserts the maximum value of 10 and the minimum as its negative without showing how these values are obtained from the Lagrange multiplier solutions. It does not compute f(x,y) at the points (x,y)=(3/2λ,4/2λ) with λ=±5/4, nor does it verify that these are indeed the extreme values. This omission could mislead a student into thinking the conclusion follows automatically from the earlier steps.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 2026-10-03 — 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 (misleading) 2026-10-03 — The final sentence asserts the maximum value of 10 and the minimum as its negative without showing how these values are obtained from the Lagrange multiplier solutions. It does not compute f(x,y) at the points (x,y)=(3/2λ,4/2λ) with λ=±5/4, nor does it verify that these are indeed the extreme values. This omission could mislead a student into thinking the conclusion follows automatically from the earlier steps.qwen3.6:27b-mlx: pass 2026-09-28 — 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: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the method of Lagrange multipliers, derives the correct critical values, and identifies the maximum and minimum values accurately.gpt-oss:20b: pass 2026-09-28
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.