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Impose the fixed conserved order parameterwith a Lagrange multiplier ${\mu }_0$. Stationarity ofunder every local variation givesThus the chemical potential is spatially constant at constrained equilibrium.

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

At a regular constrained stationary point of $F$ on $G=0$, the Lagrange multiplier method solvesFor the first problem, substitute $z=1+xy$ to obtainIts stationary equations imply either $x=y$ or $xy=0$, and in either case the only real stationary point is $x=y=0$, $z=1$. Coercivity ensures that the minimum is attained, soat $(0,0,1)$.

For the second problem, the arithmetic-geometric mean inequality givesConsequentlyEquality occurs at $(0,0,1)$, and hence

Minimize $\Vert u{\Vert }_{\Sigma }^2$ subject to $\Vert u{\Vert }_2=1$. A minimizing sequence is bounded in $\Sigma$, and part 2 supplies a subsequence converging strongly in $L^2$ and weakly in $\Sigma$. The constraint survives the strong convergence, while weak lower semicontinuity of the norm shows that the limit $\psi$ attains the minimum. Since $\Vert |\psi |{\Vert }_2=\Vert \psi {\Vert }_2$ and $|\nabla |\psi ||\le |\nabla \psi |$ almost everywhere, we may take $\psi \ge 0$.

The Lagrange multiplier equation iswhere testing with $v=\psi$ shows that $\lambda =\Vert \psi {\Vert }_{\Sigma }^2>0$. ThusThis is the ground-state eigenfunction of a confining Schrödinger operator.

For the Gibbs entropyfirst maximize subject only to ${\sum }_n{p}_n=1$ over $\Omega$ accessible microstates. A Lagrange multiplier givesso every $p_n$ is equal. Normalization yields the microcanonical ensemble

For the canonical ensemble, impose both normalization and the mean-energy constraint ${\sum }_n{p}_n{E}_n=E$. Stationarity ofgives $p_n=C{e}^{-\beta E_n}$. The canonical partition function fixes $C$, soThermodynamic consistency identifies $\beta =(k_{B}T{)}^{-1}$. This is the maximum-entropy derivation of equilibrium ensembles.

For a regular constraint $g=0$, an interior constrained extremum satisfiestogether with the constraint; boundary cases must also be checked.

Here maximize $f=xyz$ subject to $x^2+y^2+z^2=L^2$. The Lagrange multiplier equations giveAt a positive maximizer these imply $x=y=z=L/\sqrt{3}$. Boundary points have zero volume, so the maximum is

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

The expected number of failures isHolding the alternative and test size fixed, constant power is equivalent to fixing $V(n_0,n_1)=C$. The ethical allocation problem is thereforeThe Lagrange multiplier stationary equations determine the ratio; the second-order condition requires the constrained stationary point to be a local minimum.

Equilibrium minimizes $F$ at fixed total composition $\int \varphi$, so a Lagrange multiplier gives $\delta F/\delta \varphi ={\mu }_0$, independent of position. In the two homogeneous phases,The symmetric coexistence pair therefore has ${\mu }_0=0$. For a planar profile depending only on the normal coordinate $x$,orup to reversal of the two phases.

The mean extension in an ensemble of $A$ molecules iswith constraints ${\sum }_i{a}_i=A$ and ${\sum }_i{a}_i{l}_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the multinomial coefficientThus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated Lagrange multiplier expression.

Because $A\gg 1$, the Stirling formula gives $\log (a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ givesNormalization by ${\sum }_i{a}_i=A$ therefore yields the probability mass functionHere $Z$ is the fixed-tension partition function that normalizes the probabilities.

At a regular constrained extremum of $f$ on $g=0$, the tangent derivatives of $f$ vanish. Since $\nabla g$ is normal to the constraint surface, the Lagrange multiplier condition isOne solves these equations and then compares the resulting candidates, including any boundary or singular cases.

Let the base have dimensions $x$ and $y$, and let the height be $z$. Measure cardboard relative to the thickness of the front and back. The weighted amount used isbecause the bottom has triple thickness, the two front and back faces have ordinary thickness, and the two side faces have double thickness. The constraint is $xyz=3$.

The multiplier equations areThey implyThus $x=2y$ and $z=3y/2$; imposing $xyz=3$ gives $y=1$. ThereforeThis is the global minimum: by the arithmetic-geometric mean inequality,and equality holds at these dimensions. This is an instance of weighted open-box minimization.

Part b shows that $H(Q_{\alpha })$ decreases continuously from $H(Q_0)={\log }_2|\mathcal{A}|$ to $H(Q_1)=H(Q)$. Because $Q$ is nonuniform, the variance in part b is positive, so for each intermediate $R$ there is a unique ${\alpha }^{\ast }\in (0,1)$ with $H(Q_{{\alpha }^{\ast }})=R$.

To minimize $D(P\Vert Q)$ subject to $H(P)\ge R$, the optimum lies on the boundary $H(P)=R$. The Lagrange multiplier equations forgive $P(x)\propto Q(x{)}^{\alpha }$ for some $\alpha \in (0,1)$. The entropy constraint selects $\alpha ={\alpha }^{\ast }$ uniquely. Therefore

Associate a nonnegative Lagrange multiplier ${\lambda }_i$ with each inequality. The Lagrangian dual problem begins withIts infimum over $x\in {\mathbb{R}}^n$ is finite exactly when $A^T\lambda +c=0$, in which case it equals $-b^T\lambda$. The dual linear program is consequentlyFor any primal-feasible $x$ and dual-feasible $\lambda$,which proves weak duality. Strong duality means equality of the two optimal values. The stated strict feasibility is the Slater condition; together with finiteness of the primal optimum it gives strong duality and an attained dual optimum ${\lambda }^{\ast }$.

The thermodynamic order parameter is the conserved scalar composition difference $\varphi$ of the binary fluid mixture. The momentum density $\rho \mathbf{v}$ is an additional conserved hydrodynamic mode; it is not a symmetry-breaking order parameter, but it must be retained because momentum relaxes only through spatial transport.

Composition conservation gives a continuity equation,The current $\mathbf{J}=-M\nabla \mu$ is the leading isotropic, dissipative constitutive law: it drives material down gradients of the chemical potential. For the Landau-Ginzburg theorythe functional derivative isThese statements give the advective Cahn--Hilliard equation.

Constant mass density and incompressible flow require $\nabla \cdot \mathbf{v}=0$. Momentum conservation gives the Navier-Stokes equation: material acceleration equals the sum of the Newtonian viscous force $\eta {\nabla }^2\mathbf{v}$, the pressure force $-\nabla P$, and the Korteweg force $-\varphi \nabla \mu$. The pressure is the Lagrange multiplier enforcing incompressibility. Together these equations are Model H dynamics; isothermality removes the need for a separate energy equation.

The empirical kernel covariance operator $\Sigma$ is self-adjoint and positive semidefinite. Maximize $⟨u,\Sigma u⟩$ subject to $⟨u,u⟩=1$. The first variation of the Lagrange multiplier functional givesso $\Sigma v=\lambda v$. Taking the inner product with $v$ gives

In the long-wavelength $O(N)$ description of an antiferromagnet, $v$ is the antiferromagnetic spin wave velocity and $g$ controls the stiffness or strength of quantum fluctuations. The field $\lambda$ is an auxiliary Lagrange multiplier enforcing the fixed-length constraint on the order parameter. At the translation-invariant saddle, $i\lambda =m^2$ shifts every propagator denominator and $m$ is the excitation gap, which explains the name gap equation.

The Large-N expansion makes fluctuations of the auxiliary field relatively small. Its leading saddle-point condition is then self-consistent and becomes the displayed gap equation.

A phase-space action treats canonical variables as independent fields. Varying them gives Hamilton's equations; Lagrange multipliers can impose constraints.

For $\gamma >0$ and $c\ne 0$, minimize $c^{T}u+\gamma z^2$ subject to $\Vert u{\Vert }^2+z=R$. Adding a positive Lagrange multiplier times the constraint gives a positive-definite quadratic optimization Lagrangian, whose minimizer is $u=-c/(2\lambda )$ and $z=-\lambda /(2\gamma )$. Feasibility is the displayed scalar equation. Its left side strictly decreases from positive infinity to negative infinity on $\lambda >0$, so it has a unique root for every real $R$. The Lagrangian sufficiency theorem certifies a unique global optimum on the possibly nonconvex equality surface.

An optimal dual variable $y^{\ast }\in \partial p(0)$ bounds the effect of perturbing a convex value function:If $p$ is finite convex near zero, $p^{\prime }(0;d)={\max }_{y\in \partial p(0)}⟨y,d⟩$. A singleton subdifferential gives differentiability and a first-order expansion. For an upper-bound constraint relaxed by $z$, the derivative equals the negative of the nonnegative Lagrange multiplier; relaxing the bound can decrease the minimum value.

An einbein specifies a one-dimensional metric tensor along a worldline. In the relativistic particle phase-space action it is the Lagrange multiplier for the mass-shell condition and transforms as $\delta e=\dot{ϵ}$ under the corresponding canonical gauge transformation. After eliminating the momentum, its geometric interpretation depends on the normalization of the particle action.