The following section is about constrained optimization and lagrange multipliers. For more mathematical topics checkout other Awesome list.


Lagrange multipliers Introduction

The Lagrange multiplier technique lets you find the maximum or minimum of a multivariable function $f(x, y, \dots)$ when there is some constraint $\color{red}{g(x,y,\dots)}$ on the input values $x$, $y$, $\dots$ you are allowed to use.

Multivariable function $\color{blue}{f(x, y)}$ and constraint $\color{red}{g(x,y) = c}$.

the gradient of $f$ evaluated at a point $(x_0, y_0)$ always gives a vector perpendicular to the contour line passing through that point. This means when the contour lines of two functions $\color{blue}f$ and $\color{red}g$ are tangent, their gradient vectors are parallel. This tangency means their gradient vectors align:

Tangency Condition $$ \label{tangency} \nabla \color{blue}{f(x_0,y_0)} = \color{green}{\lambda_0} \nabla \color{red}{g(x_0,y_0)} $$

Here, $\color{green}{\lambda}$ represents some constant. Some authors use a negative constant, $-\color{green}{\lambda}$.

Let’s see what this looks like in our example where $\color{blue}{f(x,y)=2x+y}$ and $\color{red}{g(x,y)=x^2+y^2}$. The gradient of $f$ is

and the gradient of $g$ is

Therefore, the tangency condition from equation $\ref{tangency}$ ends up looking like this:

In the example from above, this leads to the following three equations and two three unknowns $x_0$, $y_0$ and $\lambda_0$.

Plugging $x_0$ and $y_0$ into the third equation results in a quadratic equation with two solutions for $\lambda_0$

This leads to two solutions for the input pair $(x_0,y_0)$ which can be plugged into the original function $f(x,y)$. One input leads to the minimum and the other to the maximum value of $f$.

In general the we can write these conditions by saying we are looking for constants $x_0$, $y_0$ and $\lambda_0$ that satisfy the constraint and the tangency condition \ref{tangency}.

Joseph Louis Lagrange wrote down a special new function which takes in all the same input variables as $f$ and $g$, along with the new kid in town $\lambda$, thought of now as a variable rather than a constant.

The Lagrangian $$ \mathcal{L}(x,y,\dots,\color{green}{\lambda}) = \color{blue}{f(x,y,\dots)} - \color{green}{\lambda}(\color{red}{g(x,y,\dots) - c}) $$ with the Lagrange multiplier $\color{green}{\lambda}$.


Leave a comment