다변량 극값 및 제약 최적화
도입됨| English | 의미 |
|---|---|
| Hessian/ˈheʃn/ | 헤시안 행렬 |
| Lagrange multiplier/ˈlæɡreɪndʒ ˌmʌltɪˈplaɪə/ | 라그랑주 승수 |
A decision before an answer
- A zero gradient can identify a minimum, a maximum or a saddle. It can also fail to classify a perfectly ordinary minimum.
- Your goal: Classify two-variable critical points with the Hessian.
Read the relationship
- For a differentiable function on an open domain, an interior local extremum has zero gradient. This condition is necessary, not sufficient. For two variables at a critical point, let D=f_xx f_yy−(f_xy)². If D>0, f_xx>0 gives a strict local minimum and f_xx<0 gives a strict local maximum; D<0 gives a saddle. These tests require suitable second-derivative regularity near the point.
- Use Lagrange multipliers with a regular constraint.
At a critical point f_xx=2, f_yy=−2 and f_xy=0. What does the second-derivative test give?
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D=2(−2)−0²=−4, so the point is a saddle.
Use the defining rule
- When D=0 the second-derivative test is inconclusive, not evidence of a saddle. The functions x⁴+y⁴ and x⁴−y⁴ have the same zero Hessian at the origin, but one has a strict minimum and the other changes sign. Evaluate along contrasting directions or use a direct inequality. For a quadratic form, positive definiteness gives a more general Hessian interpretation.
- Compare extrema and attainable values under regular constraints.
What is the maximum of xy subject to x²+y²=2?
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(x−y)²≥0 implies 2xy≤x²+y²=2. Equality occurs at (1,1) and (−1,−1).
Check the conditions
- On a smooth equality constraint g(x,y)=c with nonzero gradient g, solve gradient f=lambda gradient g together with g=c. These equations generate candidates; they do not classify or guarantee a global extremum. If the constraint gradient vanishes, the regularity condition fails and the multiplier equations can miss a constrained extremum. Treat such points separately. For x²+y²=a and xy=b>0, (x−y)²=a−2b shows necessity a≥2b. It is also sufficient: take s=√(a+2b), d=√(a−2b), x=(s+d)/2 and y=(s−d)/2. These real values have xy=b and x²+y²=a. A minimum condition alone needs this attainment check to establish solvability.
- Compare extrema and attainable values under regular constraints.
Minimise x²+y² subject to x+y=6. The constraint gradient is (1,1), so 2x=lambda and 2y=lambda. Thus x=y and the constraint gives x=y=3, with value 18. Completing the square on the line gives x²+(6−x)²=2(x−3)²+18, proving a global minimum. There is no maximum on this unbounded line, despite the existence of a multiplier candidate.
The minimum value of x²+y² on x+y=4 is ____.
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The minimising point is (2,2), giving 4+4=8, or complete the square.
Apply the task format
- A continuous function on a compact feasible set attains global extrema. To find them, compare all interior candidates and all boundary pieces, including corners and endpoints. For a rectangle, optimise the restrictions on each edge. For a disk, the circular boundary may use a parameter or multiplier. Solving only the unconstrained gradient ignores possible boundary winners.
- Compare extrema and attainable values under regular constraints.
A Lagrange multiplier solution is a candidate, not automatically a maximum or minimum. Check regularity and the complete feasible set.
Which answer fits this case?
Classify two-variable critical points with the Hessian
D=0 at a critical point proves that point is a saddle.
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The test is inconclusive; x⁴+y⁴ has D=0 at a strict minimum.
Keep the distinctions
- Hessian 海森矩阵 — The matrix of second partial derivatives used to study local curvature.
- Lagrange multiplier 拉格朗日乘子 — A scalar relating objective and regular constraint gradients at a constrained extremum.
- Classify two-variable critical points with the Hessian.
- Use Lagrange multipliers with a regular constraint.
- Compare extrema and attainable values under regular constraints.
Match each term with its precise meaning in this lesson.
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Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
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The order follows the stated process; check each stage before the next.