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The chosen quadratic objective function above gives more than an optimal allocation: its Lagrange multiplier predicts how much the best cost changes when the required total changes. For this example,This is an explicit derivative of a constrained value function and an instance of the envelope theorem. At $c=3$, one extra small unit of required output costs approximately $4$ units per unit of output, in the chosen normalized model. Increasing $c$ makes this equality-constrained minimization problem more expensive; it is not a gain from relaxing an upper resource limit. The sign and meaning depend on how the optimization constraint and optimization Lagrangian are written.^[ref]

For an actual finite change $\delta$, the cost difference is exactlyAt $\delta =0.1$, the linear approximation predicts $0.4$, while the exact increase is $61/150\approx 0.406667$. The extra $1/150$ comes from the squared term in this polynomial, rather than numerical error. A Lagrange multiplier gives a local sensitivity analysis; treating it as a constant price over a large change generally loses the change in the optimum itself.

For a smooth optimizer $z(c)$ of an equality problem $g(z)=c$, the short chain rule calculation explains the result: $\nabla g(z(c))\cdot z^{\prime }(c)=1$ and $\nabla f(z(c))=\lambda (c)\nabla g(z(c))$ imply $df(z(c))/dc=\lambda (c)$. The differentiability and stationarity assumptions matter: an optimizer can fail to be unique or vary smoothly. The envelope theorem records when this interpretation is justified.

UseStationarity givesPutting $t=\lambda /2$, this becomesand the constraint requiresThe left side is strictly increasing, and $t=1$ solves the equation. Hence

The objective is convex and the constraint is affine. Its tangent-plane inequality at $(1,1,1)$ gives, for every feasible $(x,y,z)$,so the Lagrange point is globally optimal. Moreover, at the dual value $\lambda =2$, the infimum of the Lagrangian function in constrained optimization is attained at the same point and equals three. The primal and dual values coincide, so strong duality holds.

For the value function, the multiplier convention above gives the derivative of a constrained value functionAt $b=6$, therefore,