Year 1 Sem 2/LC Mathematics for Physicists 1B/Lecture 06

Lecture 6: Partial Differentiation VI: Lagrange Mulipliers
Fri 30 Jan 2026 12:00

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Note: Lecture began with a continuation of 3D stationary points, which I’ve bundled under last lectures notes to not interrupt the derivation.

Lagrange Multipliers

Suppose we wanted to minimise a function $f(x, y) = x^2 + y^2$ subject to a constraint, i.e. saying $x + y = 1$ (and therefore excluding the $(0, 0)$ trivial case).

While we could do this by writing $y = 1-x$, substituting and minimising solely in terms of $x$, there is a more sophisticated method that becomes useful in more complex cases where the constraint is not invertable.

We consider the function:

$$F(x, y, \lambda) = x^2 + y^2 + \lambda(x + y - 1)$$

And minimise with respect to all three variables. When this happens, $F(x, y,\lambda) = f(x, y)$ at the minimised point.

Generally, to find the extrema of $f(x_1, x_2, \ldots, x_n)$ subject to $p$ constraints $g_i(x_1, x_2, \ldots, x_n) = 0$, where $p<n$, we introduce a new function of $n+p$ variables:

$$F(x_1, x_2, \ldots, x_n, \lambda_1, \lambda_2, \ldots, \lambda_p) = f(x_1, x_2, \ldots, x_n) + \sum_{i=1}^{p} \lambda_i g_i(x_1, x_2, \ldots, x_p)$$

We then extremise $F$ wrt all $n+p$ variables. Extremising with respect to $\lambda_i$ enforces the constraint:

$$F_{\lambda_i} = g_i(x_1, x_2, \ldots, x_n) = 0$$

So the the lambda term ends up dropping out and $F$ and $f$ have the same numerical value at extrema.

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