M408M Learning Module Pages

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Chapter 10: Parametric Equations and Polar Coordinates

Chapter 12: Vectors and the Geometry of Space


Chapter 13: Vector Functions

Chapter 14: Partial Derivatives

Learning module LM 14.1: Functions of 2 or 3 variables:

Learning module LM 14.3: Partial derivatives:

Learning module LM 14.4: Tangent planes and linear approximations:

Learning module LM 14.5: Differentiability and the chain rule:

Learning module LM 14.6: Gradients and directional derivatives:

Learning module LM 14.7: Local maxima and minima:

Learning module LM 14.8: Absolute maxima and Lagrange multipliers:

       Absolute maxima
      Lagrange multipliers I
      Lagrange multipliers II

Chapter 15: Multiple Integrals

Lagrange multipliers I

Lagrange Multipliers I On the last slide, we found the maximum of a function on the boundary of a square, and a maximum of a function on the boundary of a disk, using single-variable methods. This isn't always possible, especially when our region is 3-dimensional, so that the boundary is a 2-dimensional surface. In this slide we'll see how to find the maximum or mimimum of a function on such a curve or surface. Such so-called constrained optimization problems come up frequently in applications. Suppose, for instance, you want

     To maximize production by allocating a fixed amount of capital between labor and manufacturing costs,

     Or to know where a weather balloon at the end of a fixed length of cable ends up,

     Or need to pack food for a trip, but you've got a limited amount of space in your car.


We solve these problems with the method of Lagrange multipliers. This is a particularly powerful technique in engineering, the sciences and economics because it works in a wide variety of problems and in any number of variables. There are two ways to think about the method. One, based on gradients, is explained in the following video. The other, based on contour maps, is explained in the subsequent text.

Let's set up an idealized mathematical model by taking a 'hike in the mountains':
  Problem: Find the maximum and minimum values of $$f(x, y)\ = \ y^2 - x^2$$ subject to the constraint $$g(x,\,y)\ =\ x^2+y^2-4\ = \ 0\,.$$ We could use single variable methods: simply eliminate $y$ from $f(x,\, y)$ using $g(x,\,y) = 0$. However, it will be instructive to investigate the problem using a lot of what has been learned of late! using a lot of what has been learned of late! The graph of $x^2+y^2 - 4=0$ is a circle of radius $2$ centered at the origin in the $xy$-plane. Without restrictions on $x,\, y$ there would be no maximum or minimum values of $f(x,\,y)=x^2-y^2$, just the saddle point at the origin!


The constraint places restrictions on the values of $x,\,y$ in $f(x,\,y)$. Let's explore the effect of this constraint condition both in $3$-space and via contour maps in the $xy$-plane. In $3$-space we look at the surface $z=f(x,\,y)$ and think of $z$ as elevation above sea level. The trouble is that the graph of $f$ is a surface in $3$-space, while the graph of $g$ is a circle in the $xy$-plane. But in $3$-space the graph of $x^2+y^2 - 4 = 0$ is a circular cylinder. So the constraint $g(x,\,y)= 0$ says we look only for highest and lowest points on the path where the cylinder graph intersects the graph of $z = x^2-y^2$ as shown in orange to the left below. At these maximum and minimum points you are walking horizontally along the contour through that point - you'd still be going uphill or downhill otherwise! How can this be seen in the contour map of $z = x^2-y^2$ as shown to the right below?

Do you see which point on the orange curve corresponds to the point $P$ on the contour map? What about $Q$ and $R$? (Don't forget to rotate the surface for different viewpoints.) The crucial idea is that when the contour line and the circle have a common tangent at a point $(a,\,b)$ in the right hand graphic, then the corresponding point on the orange curve will be at a local max and local minimum on the curve because here you will be walking horizontally along the contour. But then both gradient vectors $\nabla f(a,\,b)$ and $\nabla g(a,\,b)$ will be perpendicular to this common tangent at $(a,\,b)$, hence parallel. Since two vectors are parallel when one is a scalar multiple of the other, we thus get:

Method of Lagrange Multipliers: the maximum and minimum values of $z = f(x, \,y)$ subject to the constraint $g(x, \,y) = 0$ occur at a point $(a, \,b)$ for which there exists $\lambda$ such that $$ (\nabla f)(a, \,b) \ =\ \lambda (\nabla g)(a, \,b), \qquad g(a,\, b) \ = \ 0\,,$$ and $(\nabla g)(a, \,b) \ne 0$. Such points $(a,\,b)$ will be called critical points.

The method of Lagrange multipliers works just as well when $f(x, \,y,\,z)$ and $g(x, \,y, \,z)$ are functions of $3$ variables (or any greater number of variables for that matter). Since $\nabla f$ and $\nabla g$ are still well-defined, we can still solve the equations $\nabla f = \lambda \nabla g$ and $g=0$.