 Lagrange multiplier

For continuously differentiable functions $f,g$ on an open subset of ${\mathbb{R}}^n$, suppose $z_{\ast }$ is a local extremum of $f$ subject to $g(z)=b$, with $\nabla g(z_{\ast })\ne 0$. Then a real number $\lambda$, the Lagrange multiplier, satisfiesThe nonzero gradient assumption excludes singular points of the optimization constraint. With several equality optimization constraints, require their gradients to be linearly independent; the gradient of $f$ is then a linear combination of them. These are necessary conditions, not automatic certificates of a global maximum or global minimum.

Why can the gradient be nonzero at an optimum? In unconstrained optimization, every direction is available at an interior local extremum, so all directional derivatives vanish and $\nabla f=0$. On a regular equality feasible set, only directions in its tangent space are available to first order. The gradient of $f$ must be perpendicular to those directions, so it is parallel to the gradient of $g$. In two dimensions this explains the tangency of their level sets. See MIT, Lagrange multipliers and contour geometry.

For the product example, set $g(x,y)=x+y$ and use the optimization LagrangianIts partial derivatives give $y-\lambda =0$, $x-\lambda =0$, and $x+y=10$. Hence $x=y=\lambda =5$. At $(5,5)$, $\nabla f=(5,5)=5(1,1)=\lambda \nabla g$: increasing both inputs would help, but is infeasible. Along the allowed direction $(1,-1)$, the directional derivative is zero. The endpoints must still be checked because of the nonnegativity optimization constraints; completing the square establishes the unique global maximum.

The sign convention matters: writing $L=xy+\mu (x+y-10)$ instead gives $\mu =-5$. Neither equality Lagrange multiplier has a universal nonnegativity requirement. Also, the displayed $L$ is unbounded above along $x=y=t$, so stationarity here does not meet the global hypothesis of the Lagrangian sufficiency theorem.

The Lagrange multiplier also measures the value of extra resources. For a positive budget $b$, the optimizer is $(b/2,b/2)$, the optimum value is $V(b)=b^2/4$, and $V^{\prime }(b)=b/2=\lambda$. At $b=10$, an extra $0.1$ units predicts a gain of $0.5$; the exact gain is $V(10.1)-V(10)=0.5025$. This is the envelope theorem in a computed sensitivity analysis: the multiplier translates a binding budget constraint into a marginal benefit.