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Use the optimization Lagrangian with the sensitivity sign conventionFor equality constraints the Lagrange multipliers have unrestricted signs. The Lagrangian sufficiency theorem says that if $h(x^{\ast })=b$ and, for some ${\lambda }^{\ast }$, $x^{\ast }$ globally minimizes $L_b(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes $f$ subject to the constraint. Indeed every feasible $x$ satisfiesThis proves the theorem without a convexity or differentiability assumption. Merely finding a stationary point of the optimization Lagrangian would not establish its required global minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, with value $+\infty$ when the feasible set is empty, and suppose $\varphi (b)$ is finite. The problem has the Strong Lagrangian property if a finite multiplier $\lambda$ satisfiesThis includes attainment of the dual bound by a multiplier. It does not by itself assert attainment of the primal infimum. For every multiplier, restricting the infimum to feasible $x$ gives ${\inf }_X{L}_b(x,\lambda )\le \varphi (b)$, the relevant weak duality. A non-vertical supporting hyperplane of a value function is a planewith $\varphi (u)\ge \varphi (b)+{\lambda }^T(u-b)$ for every $u$. Its vertical coefficient is nonzero and has been normalized to one; the epigraph lies above it. The two directions of the equivalence are proved in parts (a) and (b).

For the sampling allocation put $A={\sum }_i\sqrt{a_i{v}_i}$. Finite variance requires every $x_i>0$; interpret the objective at a zero coordinate as $+\infty$. An optimum uses all the budget, because scaling every positive $x_i$ up would otherwise reduce the objective while remaining feasible. The Cauchy-Schwarz inequality gives the global boundEquality requires $v_i/x_i$ to be proportional to $a_i{x}_i$. The budget then determines the unique allocation:This is the optimal cost-constrained stratified sampling allocation.

For an inequality ${\sum }_i{a}_i{x}_i\le b$, our sign convention uses $\lambda \le 0$, so $L_b=f-\lambda ({\sum }_i{a}_i{x}_i-b)$ is a lower bound on $f$ at feasible points. Its coordinate stationary equations are $-v_i/x_i^2-\lambda a_i=0$, and the displayed allocation givesThe function $v_i/x_i+(-{\lambda }^{\ast })a_i{x}_i$ has a strict global minimum at $x_i^{\ast }$: its second derivative is $2v_i/x_i^3>0$ and its value diverges at both endpoints of $(0,\infty )$. Thus the multiplier also supplies the Lagrangian sufficiency theorem certificate, with the tight inequality giving complementary slackness.

Finally, for $b+\delta b>0$,Hence the first-order change in minimal variance is $\boxed{{\lambda }^{\ast }\delta b}$. Increased resources decrease the variance. With the alternative convention $L=f+\mu ({\sum }_i{a}_i{x}_i-b)$, the multiplier is $\mu =-{\lambda }^{\ast }>0$ and the sensitivity is $-\mu \delta b$.