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The Lagrange condition finds candidates for extrema of a function subject to an equality constraint.
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Formula
λ has units of f divided by units of g; if the constraint is dimensionless, λ has the units of f.
| Symbol | Meaning | Unit |
|---|---|---|
| Objective function | unit of the objective value | |
| Constraint with g(x)=c | unit of g | |
| Lagrange multiplier | unit of f / unit of g |
\nabla f(x)=\lambda\nabla g(x)At a regular constrained extremum, the level surfaces of the objective and constraint are tangent. Their normals, the gradients, are therefore parallel: ∇f=λ∇g. This condition gives candidates and must be solved together with the constraint; boundary and singular points require separate checks.
For smooth functions and regular constraints, meaning ∇g≠0 at the candidate. Also inspect boundaries and singular points.
At a constrained extremum the objective and constraint level surfaces are tangent, so their normals are parallel.
Find a constrained extremum
For smooth functions and regular constraints, meaning ∇g≠0 at the candidate. Also inspect boundaries and singular points.
Candidates give x=±y; values are +1/2 and −1/2.
Extremize f=xy subject to x2+y2=1.
From (y,x)=λ(2x,2y), for nonzero x,y we obtain x=y or x=−y.
The constraint gives f=1/2 at x=y=±1/√2 and f=−1/2 at x=−y=±1/√2.
Check: |xy|≤(x2+y2)/2=1/2.
Forgetting the constraint after writing the gradient equation.
The equation ∇f=λ∇g must be paired with g(x)=c; only the coupled system determines candidates.
Formula cluster
Parallel gradients reduce a multivariable optimization problem to a system of equations.
Extremize f=xy subject to x2+y2=1. From (y,x)=λ(2x,2y), for nonzero x,y we obtain x=y or x=−y. The constraint gives f=1/2 at x=y=±1/√2 and f=−1/2 at x=−y=±1/√2. Check: |xy|≤(x2+y2)/2=1/2.
Resource optimization under a fixed budget or material constraint, minimum energy at a fixed volume, and extrema on curves or surfaces.
Here is how to work through a typical Lagrange multipliers (∇f = λ ∇g) task step by step:
Task
Candidates give x=±y; values are +1/2 and −1/2.
Solution path
Extremize f=xy subject to x2+y2=1.
From (y,x)=λ(2x,2y), for nonzero x,y we obtain x=y or x=−y.
The constraint gives f=1/2 at x=y=±1/√2 and f=−1/2 at x=−y=±1/√2.
Check: |xy|≤(x2+y2)/2=1/2.
Geometrically, λ scales the constraint gradient to match the objective gradient. Under suitable smoothness conditions, it also measures how the optimal objective value changes when the allowed constraint value c changes slightly. Its sign depends on how the constraint is written. In applications, λ can be read as a shadow price: locally it estimates how much the optimal objective changes when the available resource c increases slightly. This interpretation depends on whether the constraint is written as g=c or an equivalent rescaled condition.
A small feasible movement stays on the constraint. At an extremum, such a tangent movement cannot change the objective to first order. Thus ∇f is perpendicular to the constraint’s tangent space. Since ∇g is also normal to that space, the gradients must be parallel. The parallel-gradient argument considers only motions that initially satisfy the constraint. Such motions lie in the tangent space of the feasible surface. At an extremum, f has zero change along each of them, so its gradient is normal to the surface and proportional to ∇g.
Yes. The equations give stationary candidates but do not by themselves distinguish a maximum, minimum or saddle. Compare objective values at all feasible candidates and inspect boundaries on bounded regions. A constraint with ∇g=0 needs separate analysis because the multiplier condition can fail there. The condition is necessary, not sufficient by itself. Candidates can be maxima, minima or saddle points, and an open feasible region need not attain a global extremum. On a compact curve or surface, compare candidate values and separately handle points where the constraint gradient vanishes.
For independent equality constraints g1=c1 through gm=cm, the condition is ∇f=λ1∇g1+…+λm∇gm. There is one multiplier per constraint. The active constraint gradients must satisfy a regularity condition; otherwise other methods are needed. With two independent constraints, the objective gradient is a linear combination of two normal vectors. There are two multipliers and both constraint equations. The system grows accordingly; independence of the active gradients ensures a regular local feasible surface. With several constraints, solve the gradient relation together with every constraint simultaneously. A multiplier does not replace an equation defining the feasible set.
From x2+y2=1, 2|xy|≤x2+y2=1, so −1/2≤xy≤1/2. The Lagrange candidates attain both bounds. This independent inequality checks the calculations and proves that they are global extrema. The bound follows from (|x|−|y|)2≥0, giving x2+y2≥2|xy|. Equality holds exactly when |x|=|y|, at the candidates found. This independent check does more than confirm stationary points; it proves the global extrema directly. The inequality independently verifies the global bounds. In other problems, however, such an estimate does not automatically replace checking boundaries and candidates. Equal signs give the maximum +1/2, while opposite signs give the minimum −1/2.
Quanta exam set
Curated exam set for "Lagrange multipliers":
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Which formula describes Lagrange multipliers?
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How do you rearrange Lagrange multipliers for Find a constrained extremum?
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Which common mistake happens with Lagrange multipliers?
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