From what I can tell, the traditional way to teach Lagrange multipliers is to start with a function $f(x,y,z)$ and to look for extrema of $f$ subject to $g(x,y,z)=k$.

That is, we restrict $(x,y,z)$ to be on the level curve $g(x,y,z)=k$.

We then look at the level curves of $f$ and find the one(s) tangent to the level curve $g(x,y,z)=k$.

An example of this can be found here: http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

I came across what seems to me to be a different approach here: https://sites.lafayette.edu/thompsmc/files/2014/01/Section_14_8.pdf

In this pdf, the constraint $g(x,y,z)=k$ is not referred to as a level curve. Rather it's shown as a cylinder intersecting the shape in question [that is, the graph of $f(x,y,z)$]. We're looking for extrema along the curve of intersection of the 2 shapes.

We then look at level curves of both $f$ and $g$. And we find that the pair of level curves that are tangent correspond to an extrema of $f$. Unlike the traditional approach, here we have multiple level curves of $g$.

I can't seem to reconcile these two views and am wondering if there's a sense in which one of the 2 approaches outlined is a more generalized version of the other. Can someone help guide me on this?



Preliminary note. In the first tutorial the objective is a quadratic function and the constraint is a sphere. While in the second tutorial the objective is a sphere and the constraint a cylinder (or a quadratic function).

The main difference is that in the first tutorial only the variables $(x,y)$ are considered, and both $f$ and $g$ go from $\mathbb{R}^2$ to $\mathbb{R}$. In contrast, the second document depicts functions with three variables $(x,y,z)$ and represents the constraint $g(x,y)=z$ for different levels of $z$, hence the different level curves for the constraint on the last figure of page 4. If you fix $z$ to and arbitrary number, and eliminate $z$ from the choice variables, you end up with a problem which is comparable to the one considered in your first reference. Both views are consistent, but represent slightly different problems.


The examples are very similar to what my lecturer used. After the class, most students can use Lagrange multiplier method, but did not understand why they use it.

We move in space, trying to find local extrema of $f(\vec{x})$. Since we are constrained by $g(\vec{x})=k$, every infinitesimal step we take, $\vec{\delta}$, must satisfy $\nabla g(\vec{x})\cdot\vec{\delta}=0$.

$\vec{x}$ is location of a local extrema if $d(f(\vec{x}))=0$ i.e. $\nabla f(\vec{x})\cdot\vec{\delta}=0$.

Combining the two, at local extrema of $f(\vec{x})$ with constraint of $g(\vec{x})=k$, $\nabla g(\vec{x})\cdot\vec{\delta}=0$ and $\nabla f(\vec{x})\cdot\vec{\delta}=0$. Therefore, $\nabla f(\vec{x})$ is a scaled $\nabla g(\vec{x})$ i.e. $\nabla f(\vec{x})=\lambda \nabla g(\vec{x})$.

I like this explanation more as it is not limited to 3 dimensional space and it is also use a familiar concept which is first derivative is zero at local extrema.


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