Local Extrema of f(x,y) on x^2 + y^2 <= 1

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In summary: The Lagrange multiplier method will help you find extrema on the circumference of the disk. To find extrema in the disk, simply set the partial dervatives of the object function to 0 and see if you find in critical points in the disk.so in this case, where the critical points are supposed to fall within that disk, does that mean x and y will have to both be less than the radius of the disk (ie any points in the disk)?The Lagrange multiplier method will help you find
  • #1
Kuma
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Homework Statement



find the local extrema of

f(x,y) = 6x^2 -8x + 2y^2 - 5 around the closed disk x^2 +y^2 =< 1

Homework Equations


The Attempt at a Solution



I used the larangian way of doing this but not sure if its right. My solution:

F(x,y) = f(x,y) + lambda g(x,y)

where g(x,y) = x^2 + y^2 - 1

taking partials wrt to x y and lambda I get

Fx = 12x - 8 + 2x lamda

Fy = 4y + 2y lamda

Flamda = g(x,y)

setting each to 0 and solving I get

lamda = 2 from Fy

subbing into Fx I get x = 1

subbing into g(x,y) and solving I get y = ±1

so the extrema are

(1,1) (1,-1)

edit: I think I shouldve had my equation as

F(x,y) = f(x,y) - lamda g(x,y) rather than f(x,y) + lamda g(x,y)

with that equation I get

x = 0.5
y = ±sqrt 0.75
 
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  • #2
Kuma said:

Homework Statement



find the local extrema of

f(x,y) = 6x^2 -8x + 2y^2 - 5 around the closed disk x^2 +y^2 =< 1


Homework Equations





The Attempt at a Solution



I used the larangian way of doing this but not sure if its right. My solution:

F(x,y) = f(x,y) + lambda g(x,y)

where g(x,y) = x^2 + y^2 - 1

taking partials wrt to x y and lambda I get

Fx = 12x - 8 + 2x lamda

Fy = 4y + 2y lamda

Flamda = g(x,y)

setting each to 0 and solving I get

lamda = 2 from Fy

subbing into Fx I get x = 1

subbing into g(x,y) and solving I get y = ±1

so the extrema are

(1,1) (1,-1)

edit: I think I shouldve had my equation as

F(x,y) = f(x,y) - lamda g(x,y) rather than f(x,y) + lamda g(x,y)

with that equation I get

x = 0.5
y = ±sqrt 0.75

When you have an inequality constraint g <= 0, your Lagrangian should be [itex] F - \lambda g [/itex] (with [itex] \lambda \geq 0[/itex]) for a _maximization_ problem and should be [itex] F + \lambda g [/itex] (with [itex] \lambda \geq 0[/itex]) for a _minimization_ problem. Alternatively, you could use the single form [itex] F - \lambda g [/itex] for both max or min problems, but now [itex] \lambda \geq 0[/itex] for a max problem and [itex] \lambda \leq 0[/itex] for a min problem.

Basically, the way to remember this is to say that the Lagrangian should be *better than* the objective F for feasible points, so if feasible points have g <= 0 we get something bigger than F by subtracting a positive constant times g (and bigger is better when maximizing), and we get something smaller than F by adding a positive constant times g, and smaller is better when minimizing.

RGV
 
  • #3
so in this case, where the critical points are supposed to fall within that disk, does that mean x and y will have to both be less than the radius of the disk (ie any points in the disk)?
 
  • #4
The Lagrange multiplier method will help you find extrema on the circumference of the disk. To find extrema in the disk, simply set the partial dervatives of the object function to 0 and see if you find in critical points in the disk.
 
  • #5
Kuma said:
so in this case, where the critical points are supposed to fall within that disk, does that mean x and y will have to both be less than the radius of the disk (ie any points in the disk)?

I don't understand your question. In your original post you said you were confused about whether you should write F + lambga*g or F - lambda*g, as these gave different solutions. All I did was answer _that_ question; I did not say anything about x and y being less than the radius of the disk or equal to the radius, or anything else like that.

RGV
 

FAQ: Local Extrema of f(x,y) on x^2 + y^2 <= 1

What is a local extremum of a function?

A local extremum of a function is a point where the function reaches its highest or lowest value within a specific region. This means that the function is either increasing or decreasing in all directions around that point.

How do you determine the local extrema of a function?

To determine the local extrema of a function on a given region, you can use the first and second derivative tests. These tests involve finding the critical points of the function and then evaluating the second derivative at those points. A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.

What is the region in which we are looking for local extrema?

In this case, the region is defined by the inequality x^2 + y^2 <= 1. This is a circle with a radius of 1 centered at the origin. The function may have local extrema at any point within this circle.

Can a function have multiple local extrema within the same region?

Yes, a function can have multiple local extrema within the same region. This can occur when the function has multiple critical points within the region, or when the function has a saddle point, which is a point that is not a local extremum but is still a critical point.

How can local extrema be used in real-world applications?

Local extrema can be used in a variety of real-world applications, such as optimization problems in economics, physics, and engineering. They can also be used to analyze data and make predictions in fields such as data science and machine learning.

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