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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.

Capital and labour are nonnegative inputs. On the budget line, $0\le K\le b$ and $L=(b-K)/w$. Both endpoints give zero output, whereas every interior point has positive output, so the maximum is interior. The Lagrangian with the sign convention of part (i) isIts stationarity equations are $\alpha K^{\alpha -1}{L}^{\beta }=\lambda$ and $\beta K^{\alpha }{L}^{\beta -1}=w\lambda$. Dividing by the positive output gives $\alpha /K=\lambda /f$ and $\beta /L=w\lambda /f$, hence $wL=(\beta /\alpha )K$. Together with the budget constraint this yieldsTo verify the unique global maximum, maximize the logarithm of the Cobb–Douglas production function along the budget line:Its unique stationary point is therefore the unique maximum, including comparison with the zero-output boundary points.

Writing $r=\alpha +\beta$ and $C=(\alpha /r{)}^{\alpha }(\beta /(wr){)}^{\beta }$, the optimal output is $\varphi (b)=C{b}^r$. The Lagrange multiplier obtained from the first stationarity equation is $\lambda =\alpha \varphi /K_{\ast }=r\varphi /b$. Direct differentiation verifies the envelope theorem identity:For clarity, the global dual-supremum premise of part (i) need not hold for every exponent pair allowed here. If $r>1$, along $(K,L)=(t{K}_{\ast },t{L}_{\ast })$ the unrestricted Lagrangian is $t^r\varphi (b)+\lambda b(1-t)$, which tends to infinity for every finite $\lambda$. Thus the verification above is by constrained stationarity and direct differentiation, and remains valid even when that stronger global Lagrangian duality representation fails.

For fixed $b$, put ${\mathcal{L}}_b(x)=f(x)+\lambda (b{)}^T(b-g(x))$. Feasibility gives ${\mathcal{L}}_b(\bar{x}(b))=f(\bar{x}(b))=\varphi (b)$. The assumed Lagrangian duality representation says this value is the unrestricted supremum of ${\mathcal{L}}_b$, so $\bar{x}(b)$ is also an unconstrained global maximizer of this continuously differentiable function. Its first derivative vanishes:Differentiate the identity $g(\bar{x}(b))=b$ using the assumed smooth dependence of the optimizer:The chain rule now gives the envelope theoremThusEach Lagrange multiplier is the marginal change in optimal value per unit relaxation of its constraint. The proof only uses the resulting stationarity and smooth feasibility identities; it does not require differentiating the maximizer explicitly or differentiating the multiplier.