Lagrange Multiplier

Used to find the Extremum of $f(x_1, x_2, \ldots, x_n)$ subject to the constraint $g(x_1, x_2, \ldots, x_n)=C$, where $f$ and $g$ are functions with continuous first Partial Derivatives on the Open Set containing the curve $g(x_1, x_2, \ldots, x_n) = 0$, and $\nabla g \not = {\bf0}$ at any point on the curve (where $\nabla$ is the Gradient). For an Extremum to exist,

$$df = {\partial f\over \partial x_1}\,dx_1+{\partial f\over \... ...l x_2}\,dx_2+\ldots + {\partial f\over \partial x_n}\,dx_n=0.$$ (1)

But we also have

$$dg = {\partial g\over \partial x_1}\,dx_1+{\partial g\over \... ...l x_2}\,dx_2+\ldots + {\partial g\over \partial x_n}\,dx_n=0.$$ (2)

Now multiply (2) by the as yet undetermined parameter $\lambda$ and add to (1),

$$\left({{\partial f\over \partial x_1}+\lambda {\partial g\ov... ...l x_n}+\lambda {\partial g\over \partial x_n}}\right)\,dx_n=0.$$ (3)

Note that the differentials are all independent, so we can set any combination equal to 0, and the remainder must still give zero. This requires that

$${\partial f\over \partial x_k}+\lambda {\partial g\over \partial x_k}= 0$$ (4)

for all $k=1$, ..., $n$. The constant $\lambda$ is called the Lagrange multiplier. For multiple constraints, $g_1=0$, $g_2=0$, ...,

$$\nabla f = \lambda_1\nabla g_1+\lambda_2\nabla g_2+\ldots.$$ (5)

See also Kuhn-Tucker Theorem

References

Arfken, G. ``Lagrange Multipliers.'' §17.6 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 945-950, 1985.