How to Use LaGrange Multipliers to Find Highest and Lowest Points on an Ellipse?

[jemck]

It has been a while and trying to brush up on LaGrange points. I want to find the highest and lowest points on the ellipse of the intersection of the cone: x^2+y^2-z^2 ;subject to the single constraint: x+2y+3z=3 (plane).

I want to find the points and I am not concerned with the minimum and maximum yet. So far, I have done the following:

delf<x,y,z>= lambda(del)g<x,y,z>
<2x,2y.-2z>= lambda<1,2,3> where g(x,y,z)=3

Hence, 2x=1*lambda or (2x-1*lambda)=0
2y=2*lambda or (2y-2*lambda)=0
-2z=3*lambda or (-2z-3*lambda)=0

Furthermore, lambda=2x x=lambda/2
lambda=y or y=lambda
lambda=(-2/3)*z z=(-3/2)*lambda

Solving the systems I guess would be my area of problems. or putting things together: Using the constraint as a guideline, and solving for lambda, I obtain that lambda=7/3.
And, if lambda=7/3 x=7/6
y=7/3
z=-7/2

I would say that I am at the final steps of obtaining my points for the function; however, I am confusing myself with what points are permissible. ie.: (0,0,-7/2)
(0,7/3,0)
(7/6,0,0)

See what I am getting at? I would appreciate any guidance to putting this puzzle together

 

[HallsofIvy]

1) Lagrange "points" and Lagrange "multipliers" are very different things!

2) I'm not sure what you mean by " I want to find the highest and lowest points on the ellipse of the intersection of the cone". If you mean find the highest and lowest points on the ellipse formed by the intersection of the cone x²+ y²= z² and the intersection of the plane x+ 2y+ 3z= 3, then you are really trying to maximize and minimize z subject to the two constraints.

For what you give, find max and min of x²+ y²- z² subject to the constraint x+ 2y+ 3z= 3, what you've done at the beginning is exactly right: If f(x,y,z)= x²+ y²- z², then del f= 2x i+ 2yj- 2zk. We can write the constraint as g(x,y,z)= x+ 2y+ 3z= 3 so del g= i+ 2j+ 3k is perpendicular to the plane. We must have del f= 2xi+ 2yj- 2zk= &lambda;(i+ 2j+ 3k) or 2x= &lamda;, 2y= 2&lamda;, -2z= 3&lambda;.
I don't see how you got &lambda;= 7/3! You correctly observe that x= &lamda;/2, y= &lambda;, z= -(3/2)&lambda; so x+ 2y+ 3z= (1/2)&lambda;+ 2&lambda;- (9/2)&lambda;= (1/2+ 4/2- 9/2)&lamda;= -2&lamda;= 3: &lambda= -3/2. From that,
x= &lamda;/2= -3/4, y= &lambda;= -3/2, and z= (-3/2)&lambda;= 9/4.

Myself, I would solve in a slightly different way: since we don't really need to find &lambda; itself, it is simplest to eliminate &lambda; by dividing equations: The second equation divided by the first gives 2y/2x= 2&lamda;/&lambda; or y/x= 2 so y= 2x. The third equation divided by the first gives -2z/2x= 3&lamda;/&lamda; or -z/x= 3 so z= -3x. Now put those into the constraint: x+ 2y+ 3z= x+ 2(2x)+ 3(-3x)= -4x= 3. x= -3/4, y= 2x= -3/2, z= -3x= 9/4, just as before.

Those values "go together", you don't "mix and match" with 0! The one extremal point is (-3/4, -3/2, 9/4).

If the problem really is "find extermal values of z subject to both x²+ y²- z²= 0 and x+ 2y+ 3z= 3", do it as a "double multiplier problem: del z= k must be a linear combination of del (x²+ y²- z²) and del (x+ 2y+ 3z). That is, k= &lambda;₁(2xi+ 2yj- 2zk)+ &lambda;₂(i+ 2j+ 3k) or 2&lambda;₁x+ 2&lambda;₂= 0, 2&lambda;₁y+ 2&lambda;₂= 0, -2&lambda;₁z+ 3&lambda;₂= 1. Together with the two constraints that's 5 equations for 5 unknowns.

 

[M98Ranger]

.It looks like you are on the right track in solving for the Lagrange multipliers, but there are a few things to keep in mind when finding the points of interest on the ellipse.

First, it is important to note that the Lagrange multiplier method will only give you the critical points on the ellipse, which may not necessarily be the highest and lowest points. To find the maximum and minimum points, you will need to evaluate the function at these critical points and compare the values.

Second, when solving for the Lagrange multipliers, you will typically have multiple solutions, as you have found. However, not all of these solutions will be valid points on the ellipse. In this case, you can use the constraint equation to eliminate some of the solutions that do not satisfy the constraint.

For example, using the constraint x+2y+3z=3, we can eliminate the solution (0,0,-7/2) since it does not satisfy the constraint. This leaves us with the two solutions (0,7/3,0) and (7/6,0,0). These two points can then be evaluated to determine which is the maximum and which is the minimum.

In summary, to find the highest and lowest points on the ellipse of the intersection of the cone and plane, you will need to solve for the Lagrange multipliers, use the constraint equation to eliminate any invalid solutions, and then evaluate the remaining points to determine which is the maximum and which is the minimum. I hope this helps clarify the process for you. Good luck with your problem!