Lagrange multipliers
Problem 10.116 · medium
Use Lagrange multipliers to find the maximum and minimum of \( \displaystyle f(x, y) = 2 x + 2 y \) on the circle \( \displaystyle x^2 + y^2 = 16 \).
- Solve ∇f = λ∇g with g(x, y) = x² + y² − r² = 0.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(2 x + 2 y\right)\\\frac{\partial}{\partial y} \left(2 x + 2 y\right)\end{matrix}\right] = \left[\begin{matrix}2\\2\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + y^{2} - 16\right)\\\frac{\partial}{\partial y} \left(x^{2} + y^{2} - 16\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\2 y\end{matrix}\right] \]∇g.✓ Proved
- So x = 2/(2λ), y = 2/(2λ); substituting into the constraint gives λ = ±√(8)/(2√16).
- \[ 8 \sqrt{2} \]The maximum; the minimum is its negative.✓ Proved
Answer \( \max = 8 \sqrt{2},\ \min = - 8 \sqrt{2} \)
Lines: 3 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 | ✓ 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 | f around 100,000 points of the circle tops out at the same value |
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 2026-09-26 with SymPy 1.14.0.