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 II

Lagrange Multipliers II

In the last slide we derived the formulas for the
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 λ such that (f)(a,b) = λ(g)(a,b),g(a,b) = 0,
and (g)(a,b)0.

It doesn't matter whether you think of them in terms of gradients or contour lines, since the formulas, and the method, are the same. Now let's use the method of Lagrange multipliers to solve a couple of problems, starting with a slightly more complicated version of our hiking problem:

  Example 1: Use Lagrange multipliers to find the minimum value of f(x,y) = 2x2+y2+3
subject to the constraint g(x,y) = x2+4y24 = 0.


Solution: The minimum value of f(x,y) = 2x2+y2+3
subject to the constraint g(x,y) = x2+4y24 = 0
occurs at solutions of f(x,y)=λ(g)(x,y),  g(x,y)=0.
Now f(x,y)=4xi+2yj,
g(x,y)=2xi+8yj.
thus the critical points occur at solutions of 4x=2λx,  2y=8λy,  x2+4y24=0,
i.e., x2+4y24=0 and 2x(2λ)=0,2y(14λ)=0.
So x=0 or λ=2. Now
  (i) if x=0, x2+4y24=0y=±1;
while
  (ii) if λ=2, 2y(14λ)=0y=0,
and then x2+4y24=0x=±2.
So the critical points are (0,1),(0,1),(2,0),(2,0).
But f(0,1) = f(0,1)= 4,
f(2,0) = f(2,0) = 11.
Consequently, on g(x,y)=0 the minimum value of f(x,y) is 4, and it occurs at (0,±1).


    A standard application of Lagrange Multipliers occurs in maximizing production of 'widgets' by allocating a fixed amount of capital between labor and manufacturing costs. It is based on so-called Cobb-Douglas functions z=Cxayb relating labor, capital and output in an industrial economy as first derived by economist Paul Douglas and mathematician Charles Cobb long before financial engineering became a 'hot topic'.

  Example 2: By investing x units of labor and y units capital, Texas Tees can produce P(x,y) = 40x3/5y2/5
T-shirts. Determine the maximum number of T-shirts that can be produced on a budget of $10,000 if labor costs $100 per unit and capital costs $200 per unit.


Solution: We have to maximize the function P(x,y) = 40x3/5y2/5
subject to the budget constraint g(x,y) = 100x+200y10,000 = 0.
This maximum occurs at solutions of P(x,y) = λ(g)(x,y),g(x,y) = 0.
Now Px = 40(35x2/5y2/5),
Py = 40(25x3/5y3/5).
Thus P(x,y)=8x2/5y3/5(3yi+2xj),
while g(x,y)=100i+200j.
So the critical points occur at solutions of (8x2/5y3/5)3y = 100λ,
(8x2/5y3/5)2x = 200λ,
which simplifies to λ8x2/5y3/5 = 3y100 = 2x200,
i.e., y=13x.
  Substituting for y=13x in the budget constraint condition, we find that x = 60,y = 20.
Since P(60,20) = 40(60)3/5(20)2/5  1546.55,
the maximum number of T-shirts that can be produced is 1546.


We work two additional problems in the following video:

  1. Find the local maxima and minima of the function f(x,y,z)=x2+2y23z2 on the sphere x2+y2+z2=1.
  2. In the xy plane, find the point(s) on the ellipse (x+1)29+y2=1 that are closest to the origin.