= Lagrange multiplier as a marginal cost
{c}
{title2=$V'(c)=\lambda_*(c)$}
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,
$$
V(c)=\frac{2c^2}{3},\qquad V'(c)=\frac{4c}{3}=\lambda_*(c).
$$
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.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§5.6.3, printed pp. 251–253: differentiable optimal value and multiplier sensitivity; the book's plus-sign equality convention gives the opposite multiplier sign]{ref}
For an actual finite change $\delta$, the cost difference is exactly
$$
V(3+\delta)-V(3)=4\delta+\frac23\delta^2.
$$
At $\delta=0.1$, the <linear approximation> predicts $0.4$, while the exact increase is $61/150\approx0.406667$. 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.
\Image[/lagrange-value-function-marginal-cost.png]
{title=The multiplier is the slope of the optimal cost, with a computable finite-change correction}
{description=Original calculations for $V(c)=2c^2/3$. Left: the <value function> and its tangent at $c=3$. Right: exact and first-order cost increases for nonnegative changes $\delta$; their gap is $2\delta^2/3$. The marked $\delta=0.1$ is a chosen example, not an observed price or a measured response.}
For a smooth optimizer $z(c)$ of an equality problem $g(z)=c$, the short <chain rule> calculation explains the result: $\nabla g(z(c))\cdot z'(c)=1$ and $\nabla f(z(c))=\lambda(c)\nabla g(z(c))$ imply $d f(z(c))/dc=\lambda(c)$. 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.
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