Extrema/LaGrange in Vector Calc

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In summary, the conversation discusses finding the extrema of a function subject to a constraint. The attempt at a solution involves using the equations \nablaf and \nablag to find the extrema, but it is determined that there are no extrema for this set of equations. The constraint x-y=2/(x+y) is unbounded from both above and below, meaning there are no extrema for the given function.
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DougD720
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Homework Statement



Find the extrema of f(x,y)=x-y ; subject to x2-y2=2


Homework Equations



[tex]\nabla[/tex]f=[tex]\lambda[/tex][tex]\nabla[/tex]g


The Attempt at a Solution



[tex]\nabla[/tex]f=(1,-1)
[tex]\nabla[/tex]g=(2x, -2x)

(1,-1)=[tex]\lambda[/tex](2x, -2x)

1 = [tex]\lambda[/tex](2x) -> [tex]\lambda[/tex]=[tex]\frac{1}{2x}[/tex]

-1 = [tex]\lambda[/tex](-2y) -> [tex]\lambda[/tex]=[tex]\frac{1}{2y}[/tex]

Which means x = y , but it has to satisfy x2-y2=2 and if x=y then it cannot satisfy this meaning there are no extrema for this set of equations.

Am i right? I tried working it out with other methods but it just keeps not working, however, i plotted the two equations in 3D on Maple and they do intersect so shouldn't there be extrema? Or is the fact that x-y is a plane parallel to the xy-axis mean that all points are extrema?

We never did a problem like this in class, one with no apparent solution, so I'm confused a bit here.

And i just did another problem where I'm coming up with a solution that doesn't satisfy one of the constraints... ugh... what am i doing wrong?

Thanks for the help!
 
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  • #2
Yes, you aren't doing anything wrong. Your constraint says (x-y)=2/(x+y). So (x-y) is unbounded from both above and below. There are no extrema.
 
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1. What is an extremum in vector calculus?

An extremum, also known as an extreme value, is a point where a function has a maximum (maximum extremum) or minimum (minimum extremum) value. In vector calculus, extremum refers to the maximum or minimum value of a vector function in a given domain.

2. What is the use of the LaGrange multiplier in vector calculus?

The LaGrange multiplier is a mathematical tool used to find the maximum or minimum value of a function subject to constraints. In vector calculus, it is used to optimize vector functions with constraints by incorporating the constraints into the objective function.

3. What is the difference between a local and global extremum in vector calculus?

A local extremum is a point where a function has a maximum or minimum value in a specific region, while a global extremum is the maximum or minimum value of a function in its entire domain. In vector calculus, finding a global extremum involves checking for local extrema in all possible regions.

4. How do you find the LaGrange multipliers in vector calculus?

To find the LaGrange multipliers in vector calculus, you first set up the Lagrangian equation, which is the objective function plus the constraints multiplied by the corresponding Lagrange multipliers. Then, you take partial derivatives of the Lagrangian with respect to each variable and set them equal to zero to solve for the LaGrange multipliers.

5. What are some real-life applications of Extrema/LaGrange in vector calculus?

Extrema and LaGrange are widely used in various fields, including economics, physics, engineering, and optimization problems. For example, in economics, they can be used to determine the optimal production level for a company. In physics, they can be used to find the path of a particle with the shortest travel time. In engineering, they can be used to optimize the design of structures or systems.

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