Lagrange multipliers elliptic paraboloid

In summary, the conversation discusses using Lagrange multipliers to find the highest and lowest points on the curve of intersection between the elliptic paraboloid z=x^2+4y^2 and the right circular cylinder x^2+y^2=1. The object function is z, with the conditions that z=x^2+4y^2 and x^2+y^2=1. By solving the 5 equations using Lagrange multipliers, the maximum points are found at (-1,0,1), (1,0,1), (0,-1,1), and (0,1,1), while the minimum points are found at (-1,0,1) and (1,0,
  • #1
iceman
Hi, I'm really stuck on this problem and I need some help??

Here's the question:

The intersection of the elliptic paraboloid z=x^2+4y^2 and the right circular cylinder x^2+y^2=1. Use Lagrange multipliers to find the highest and lowest points on the curve of intersection.

Your help will be very much appreciated.
 
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  • #2
By "highest and lowest points", I take it you mean maximize and minimze the z coordinate?

In that case your "object function" is z and you have the conditions that z=x^2+4y^2 (or x^2+ y^2- z= 0) and x^2+y^2=1.

Okay, look at F(x,y,z)= z+ lambda(x^2+ 4y^2- z)+ mu(x^2+y^2-1)
(lambda and mu are the Lagrange multipliers).

The gradient of F is
grad F= (2 lambda x+ 2 mu x)i+ (8 lamda y+ 2 mu y)j+ (1- lambda)k

So we have the 5 equations
2 lambda x+ 2 mu x= 0
8 lambda y+ 2 mu y= 0
1- lambda= 0 (which kinda tells you what lamda is!)
x^2+ 4y^2- z= 0 and
x^2+ y^2= 1
to solve for the 5 unknowns x, y, z, lambda, mu.

I get that the maxima z occur at (-1,0,1), (1,0,1), (0, -1,1)
and (0,l,1).

Are you required to use Lagrange multipliers?

It simpler to rewrite z= x^2+ 4y^2= x^2+y^2+ 3y^2= 3y^2+ 1 and maximize that and to rewrite z= x^2+ 4(1-x^2)= -3x^2+ 4 and minimize that.
 
  • #3
Hi HallsofIvy, thanks for helping me with my problem. I have a couple of questions I would like to ask you about it.

1) Could you please explain step by step how you solved the 5 equations?
2 lambda x+ 2 mu x= 0
8 lambda y+ 2 mu y= 0
1- lambda= 0 (which kinda tells you what lamda is!)
x^2+ 4y^2- z= 0 and
x^2+ y^2= 1
to solve for the 5 unknowns x, y, z, lambda, mu.

As I found it difficult to decipher what the values of x, y, z, lamda, and mu were?

2) You got that the maxima z occur at (-1,0,1), (1,0,1), (0, -1,1) and (0,l,1).

I'm assuming that (0,l,1) is a typo and is meant to be (0,1,1).

As you have only found the maxima points, the above solution...
How would you use lagrange to find the lowest (minima) points on the curve of intersection?

Your help is much appreciated.
Regards
Iceman
 
  • #4
Actually, I miswrote when I said "maximum". Obviously, y= 0 is a minimum of the function z= 1+ 3y^2.

By letting y^2= 1- x^2, in x^2+ 4y^2= z, we get
z= x^2+ 4(1- x^2)= 4- 3x^2. Then z'= -6x so x= 0 is a maximum for that function. If x= 0, then, from x^2+ y^2= 1, y= 1 or -1 and
z= 0^2+ 4(1)= 4 is the maximum.


Now, solving the equations from the Lagrange multiplier:

From "lambda- 1= 0" we get lambda= 1.

Then "2 lambda x+ 2 mu x= 0" become 2x+ 2mu x= 2x(1+ mu)= 0 so either x= 0 or mu= -1.

If x= 0 then the equation x^2+ y^2= 1 becomes y^2= 1 so y= 1 or -1.
If x= 0 and y= 1 or -1 then x^2+ 4y^2- z= 0 gives z= 4.

If x is not 0, then mu= -1 so 8 lambda y+ 2 mu y= 0 becomes
8y+ -2y= 0 or y= 0. With y= 0, x^2+ y^2= 1 gives x^2= 1 and x= 1 or x= -1. With x= 1 or -1, y= 0, z= 1.

That is, (-1,0,1) and (1,0,1) give the MINIMUM values of z while

(0,1,4) and (0,-1,4) give the maximum values of z.
 
  • #5
Hey cheers for your help HallsofIvy, I appreciate it.
 

What is a Lagrange multiplier?

A Lagrange multiplier is a mathematical tool used in optimization problems to find the maximum or minimum value of a function subject to constraints.

How does a Lagrange multiplier work?

A Lagrange multiplier works by introducing a new variable, called the multiplier, into the original function and its constraints. The new function is then solved using partial derivatives to find the optimal value.

What is an elliptic paraboloid?

An elliptic paraboloid is a three-dimensional shape that resembles a bowl or a saddle. It is defined by the equation z = x^2 + y^2, where the x and y values create a parabola and the z value increases or decreases along the parabola's curve.

How are Lagrange multipliers used with an elliptic paraboloid?

Lagrange multipliers can be used to find the highest or lowest point on an elliptic paraboloid. The function to be optimized would be the equation of the paraboloid, and the constraints would be the values of x and y.

What are some real-world applications of Lagrange multipliers and elliptic paraboloids?

Lagrange multipliers and elliptic paraboloids are used in various fields, such as economics, engineering, and physics, to optimize and model complex systems. For example, they can be used to find the minimum cost of producing a product or the maximum efficiency of a machine.

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