Multivariable extrema and constrained optimisation
| English | Português |
|---|---|
| Hessian/ˈheʃn/ | Hessian |
| Lagrange multiplier/ˈlæɡreɪndʒ ˌmʌltɪˈplaɪə/ | Lagrange multiplier |
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?
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?
(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 ____.
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.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.