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+---
+title: "Lagrange multiplier"
+firstLetter: "L"
+publishDate: 2021-03-02
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-02T16:28:42+01:00
+draft: false
+markup: pandoc
+---
+
+# Lagrange multiplier
+
+The method of **Lagrange multipliers** or **undetermined multipliers**
+is a technique for optimizing (i.e. finding the extrema of)
+a function $f(x, y, z)$,
+subject to a given constraint $\phi(x, y, z) = C$,
+where $C$ is a constant.
+
+If we ignore the constraint $\phi$,
+optimizing $f$ simply comes down to finding stationary points:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
+\end{aligned}$$
+
+This problem is easy:
+$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary,
+so all we need to do is find the roots of
+the partial derivatives $f_x$, $f_y$ and $f_z$,
+which we respectively call $x_0$, $y_0$ and $z_0$,
+and then the extremum is simply $(x_0, y_0, z_0)$.
+
+But the constraint $\phi$, over which we have no control,
+adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$,
+so if two are known, the third is given by $\phi = C$.
+The problem is then a system of equations:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
+    \\
+    0 &= \dd{\phi} = \phi_x \dd{x} + \phi_y \dd{y} + \phi_z \dd{z}
+\end{aligned}$$
+
+Solving this directly would be a delicate balancing act
+of all the partial derivatives.
+
+To help us solve this, we introduce a "dummy" parameter $\lambda$,
+the so-called **Lagrange multiplier**, and contruct a new function $L$ given by:
+
+$$\begin{aligned}
+    L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z)
+\end{aligned}$$
+
+Clearly, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$,
+so now the problem is a single equation again:
+
+$$\begin{aligned}
+    0 = \dd{L}
+    = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z}
+\end{aligned}$$
+
+Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$.
+This choice represents satisfying the constraint,
+so now the remaining $\dd{x}$ and $\dd{y}$ are independent again,
+and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$.
+
+This generalizes nicely to multiple constraints or more variables:
+suppose that we want to find the extrema of $f(x_1, ..., x_N)$
+subject to $M < N$ conditions:
+
+$$\begin{aligned}
+    \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M
+\end{aligned}$$
+
+This once again turns into a delicate system of $M+1$ equations to solve:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N}
+    \\
+    0 &= \dd{\phi_1} = \phi_{1, x_1} \dd{x_1} + ... + \phi_{1, x_N} \dd{x_N}
+    \\
+    &\vdots
+    \\
+    0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N}
+\end{aligned}$$
+
+Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$
+and define $L(x_1, ..., x_N)$:
+
+$$\begin{aligned}
+    L = f + \sum_{m = 1}^M \lambda_m \phi_m
+\end{aligned}$$
+
+As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$
+to eliminate $M$ of its $N$ terms:
+
+$$\begin{aligned}
+    0 = \dd{L}
+    = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n}
+\end{aligned}$$