# How to explain lagrange multipliers to a lay audience?

So I will be giving a seminar to a scientifically mature lay audience (think bio/social science undergrad level). I have been told that I should count on less than half the audience to have experience with calculus. I think I can explain the basic optimization of $f(x)$ pictorially and introducing (and for some re-introducing) the concept of a derivative. Then I plan to go on to Lagrange multipliers and discrete time optimal control. I don't want to lose them though, so I want to know if anyone has any good explanations, analogies, videos, graphics (or anything else) that will keep a lay audience engaged. Note I will use Lagrange multipliers in the talk to explain a particular application, but I want to convey the message that the math is interesting in and of itself and that it can be used for a variety of things. I'm less concerned about the details being correct than getting the main message across. This portion of the seminar should be less than 15 minutes so I don't have that much time.

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If you only need to convey the basic intuition about the use of Lagrange multipliers in optimization and you are really worried about keeping your audience engaged, you might want to rely on their every-day intuition about level curves using this kind of drawing

(here is the .svg of the picture if you want to generate variants of it. You should probably edit it with Latexdraw 2.0 if you want to be able to alter the path, the gradient, or anything which is not in the background)

• You start by giving them the intuition that "you are at a critical point only if the path you are walking on does not cross the level curve, but rather is tangent to it".
• Then you explain how this can be checked using gradients.
• Finally, you try to make them understand that this is equivalent to the existence of a Lagrange multiplier.

People might be a little puzzled by the last step. But you can hope that most of them will catch the two first steps. Even if they don't get the last one, they may remember that the existence of the multiplier is a convenient way to check that we are at a critical point based on the algebraic formulation of the function.

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I really like this idea of using level curves! This is great. – MHH Feb 23 '14 at 4:22

Here's a terse attempt to convey the main mathematical idea geometrically.

If $f$ is a function of several independent variables, there is an associated gradient vector field $\nabla f$, made up of the partial derivatives of $f$, that has the following property: If $x_{0}$ is a point of the domain and $v$ is a unit vector, the directional derivative $$\frac{d}{dt}\bigg|_{t=0} f(x_{0} + tv) = \nabla f(x_{0}) \cdot v$$ measures the rate of change of $f$ at $x_{0}$ in the direction $v$.

If $\nabla f(x_{0})$ is non-zero, and if $v$ makes angle $\theta$ with the gradient, the directional derivative reduces to $\|\nabla f(x_{0})\| \cos\theta$. In particular, traveling along the gradient field ($\theta = 0$) causes $f$ to increase "as rapidly as possible", and traveling orthogonal to the gradient $(\theta = \pi/2$) keeps $f$ constant to first order. Geometrically, the gradient field is orthogonal to the level sets of $f$.

Now, suppose the variables are subject to a smooth constraint of the form $g(x) = c$, and we want to find local extrema of $f$. A necessary condition is that $\nabla f$ be orthogonal to the constraint, namely, that $\nabla f$ be parallel to $\nabla g$, or that $\nabla f = \lambda \nabla g$ for some scalar $\lambda$. Indeed, if this is not the case, then the projection of $\nabla f$ onto the tangent space of the constraint set gives a direction in which $f$ increases.

The diagram shows a linear function $f(x, y) = ax + by$ subject to a constraint $x^{2} + y^{2} = c$. Here $\nabla f = (a, b)$ is constant, $\nabla g = (2x, 2y)$, and the constrained extrema of $f$ occur at the points where $(a, b)$ is perpendicular to the circle.

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I think you're going to lose them at "gradient vector field". Maybe even "several independent variables". Most of the audience has never seen calculus. – user2357112 Feb 23 '14 at 3:56
This is great, but it's more geared towards a freshman level math major as opposed to a lay audience. – MHH Feb 23 '14 at 4:24
My post was an exercise in paring away technicalities while leaving a "mathematically honest" skeleton. What actually gets said in the talk would naturally require additional translation (or omission of detail) tailored to the audience's background and the intended application. It was certainly not a proposed transcript. :) But I do think there's a "lower bound" for the audience's ability to gain insight from an explanation; if this bound is not met, the talk might better omit Lagrange multipliers altogether. The bottom line is: What insights do you (MHH) expect the audience to take away? – Andrew D. Hwang Feb 23 '14 at 12:31