Sec2Sess40

From Calculus Notes
Jump to navigation Jump to search

Unformatted notes below. Notes on my shorthand method

Session 40: Proof of Lagrange Multipliers

Why are Lagrange Multipliers valid?

At constrained min/max, in any direction along the level set g=c the rate of change of f must be 0.

For any <math>\hat{u}</math> tangent to g=c (constraint) we must have df/ds(<math>\hat{u}</math>) = 0 = grad(f)*<math>\hat{u}</math> where df/ds(<math>\hat{u}</math>) is df/ds in the direction of <math>\hat{u}</math>

So any such <math>\hat{u}</math> is perpendicular to grad(f) because grad(f) is perp. to the level set of g

Another vector is perp. to the level set of g, that vector is grad(g). Therefore grad(f) is parallel to grad(g).

Warning: method does not tell if a solution is min or max

The second derivative test is not able to be used to find mix or max also.

Values can be compared to investigate if critical points are min or max. Look at solutions and see which ones produce the highest and lowest values.

To find min (or max) we compare values of f at the various solutions to Lagrange equations. In some cases the answers can be found by thinking geometrically, looking for which points are closest to the origin.

Description of finding critical points: for the function w = f(x,y,z) constrained by g(x,y,z) = c (c a constant) the critical points are defined as those points, which satisfy the constraint and where grad(f) is parallel to grad(g). In equations:

grad(f)(x,y,z) = lamda*grad(g)(x,y,z) and g(x,y,z) = c.