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Mathematics II

Lagrange Multipliers

Business I 292 words Free to read

Optimising Under Constraint

Most economic decisions are constrained: maximise utility subject to a budget or cost. Lagrange multipliers solve these problems systematically.

The setup: maximise f(x,y)f(x,y) subject to g(x,y)=cg(x,y) = c. Form the Lagrangian:

L(x,y,λ)=f(x,y)λ(g(x,y)c)\mathcal{L}(x,y,\lambda) = f(x,y) - \lambda \bigl(g(x,y) - c\bigr)

First-order conditions require setting all partial derivatives to zero:

DerivativeConditionMeaning
L/x\partial \mathcal{L} / \partial xfxλgx=0f_x - \lambda g_x = 0Marginal return per cost unit
L/y\partial \mathcal{L} / \partial yfyλgy=0f_y - \lambda g_y = 0Equalised across variables
L/λ\partial \mathcal{L} / \partial \lambdag(x,y)=cg(x,y) = cEnforces the constraint

The shadow price λ\lambda is the rate at which ff changes if the constraint relaxes by one unit: λΔfΔc\lambda \approx \frac{\Delta f^*}{\Delta c}. Geometrically, level curves are tangent here.

The Shadow Price in Action

The multiplier λ\lambda answers how much a constraint costs. In utility maximisation, the tangency condition is:

MUxpx=MUypy=λ\frac{MU_x}{p_x} = \frac{MU_y}{p_y} = \lambda

This equalises marginal utility per euro spent. If λ\lambda is large, the constraint is tight; if near zero, it is slack.

Common pitfall: Paying more than λ\lambda to relax a constraint. If overtime costs 40 euro/hour but λ=25\lambda = 25 euro, expanding destroys value.

The Lagrange recipe:

StepAction
1Write L=fλ(gc)\mathcal{L} = f - \lambda(g - c)
2Set fx=λgxf_x = \lambda g_x and fy=λgyf_y = \lambda g_y
3Set L/λ=0\partial\mathcal{L}/\partial\lambda = 0
4Solve the system; read λ\lambda as the bonus

Always report λ\lambda alongside the optimum: it prices the constraint and drives business decisions.

The Shadow Price

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Mathematics II