I don't understand the point of lagrange multipliers?

Question

I'm looking at the solutions and I don't understand why they're solving it in a way they're solving it. I'm running out of patience, please help. What is the point of lagrange multipliers?

Answer 1

I will add to your problems. Suppose you want to solve a problem with certain conditions ( constraints), then you use the method Lagrange multiplier. For example perimeter is fixed and you want to know which is the that has a maximum area, you follow this method. It gives both the maximum and minimum. You have to select the one you want. Example. A rectangle has side x and y. The perimeter is fixed to be a. Then you have condition

Area = g(x,y) =xy and perimeter 2x+2y=a

Now you have to find x and y. So define a new function

h(x,y) = g(x,y) - L ( 2x+2y-a)

L is the Lagrange undetermined multiplier.

Now take partial derivative of h(x,y) wrt x and y and equate them to 0, separately. You get

y- 2L=0, x- 2L=0

Use this in 2x+ 2y= a= 4L+4L=8L

x= a/4, y=a/4. This shows that a square has a maximum area. This method is extensively in Theoretical Physics. The word used is the extremum. But in Physics it is assumed to be a minimum. Whole classical and quantum mechanical analysis are based on this principle.

Answer 2

The point of Lagrange multipliers is constraints. Clearly you are seeing problems like z = x^2 +5y, find the maximum of z on the circle 25 = x^2 + y^2.



Why not just substitute x^2 = 25 - y^2 and be done with it you say? Certainly. Why not? But lagrange multipliers are not used to solve toy problems like this, in the words of wikipedia:



"One of the most common problems in calculus is that of finding maxima or minima (in general, "extrema") of a function, but it is often difficult to find a closed form for the function being extremized. Such difficulties often arise when one wishes to maximize or minimize a function subject to fixed outside equality constraints. The method of Lagrange multipliers is a powerful tool for solving this class of problems without the need to explicitly solve the conditions and use them to eliminate extra variables."



A simple case of a "non closed form" is an integral, as in the case of an action using "calculus of variations". You don't even know what the function is--you're trying to find it. In these cases, a trivial substitution is useless/nonexistent, but multipliers still work.



https://ocw.mit.edu/courses/mathematics/18-086-mathematical-methods-for-engineers-ii-spring-2006/readings/am72.pdf

Answer 3

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