My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 5th, 2011, 02:51 PM   #1
Newbie
 
Joined: Mar 2011

Posts: 22
Thanks: 0

Extremum search

Hello,
I have the following exercise:


I want to search the extremum with help of a matrix.


So I get the following matrix:


In this example is it a local minimum because of -2.
But I dont understand this very well, can someone explain this to me.

What I have to to is take the derivatives like above, fill in (0,0) and look at the numbers that come out, ok I got that
BUT wy can you say it is a local minimum because of the -2?

In another exercise I have (longer) I get the following matrix:


Because of eigenvalue 2 we have a local minimum.

If you want I will post the entire exercise but because of the length I didn't post it in the first place, just ask.

Thank you
thescratchy is offline  
 
March 5th, 2011, 04:32 PM   #2
Newbie
 
Joined: Mar 2011

Posts: 22
Thanks: 0

Re: Extremum search

Anyone an idea?

Or didn't I explain my question good enough?
thescratchy is offline  
March 5th, 2011, 05:46 PM   #3
Senior Member
 
Joined: Feb 2009
From: Adelaide, Australia

Posts: 1,519
Thanks: 3

Re: Extremum search

f(x,y) = x-2xy+2y

Check your work.

but the matrix is correct and the partial derivatives exist everywhere.

For a critical point, you need to solve 2x-2y=0 and -2x+4y=0; these are both lines and they intersect at (0,0).
The determinant of the matrix is (2)(4)-(-2)(-2) = 8-4 = 4, which is positive, therefore the fact that is positive shows that (0,0) is a minimum.

Also, the reason you don't get an immediate reply is that this is a forum rather than a chatroom.
aswoods is offline  
March 5th, 2011, 06:05 PM   #4
Newbie
 
Joined: Mar 2011

Posts: 22
Thanks: 0

Re: Extremum search

The error you noticed was due to copying, Thank you for noticing.

So what you are saying is:

To check if it is a critical point, I have to solve: partials derivatives = 0

If they intersect at (0,0) check:
If the determinant of the matrix is positive it is a minimum, else if the determinant is negative it is a positive.

but what if the partials derivatives intersect at minus something, or plus something?
thescratchy is offline  
March 5th, 2011, 06:46 PM   #5
Newbie
 
Joined: Mar 2011

Posts: 22
Thanks: 0

Re: Extremum search

Do you fill in the matrix the values found by solving the derivates = 0 if needed?
thescratchy is offline  
March 5th, 2011, 07:27 PM   #6
Senior Member
 
Joined: Feb 2009
From: Adelaide, Australia

Posts: 1,519
Thanks: 3

Re: Extremum search

The function f needs to have partial derivatives everywhere in the region.

Then, critical points are points where .

Second partial derivative test
If the determinant is positive, then the critical point is a maximum (if ) or a minimum (if ).
If the determinant is negative, the critical point is a saddle point.
If the determinant is zero, this test fails.
aswoods is offline  
March 5th, 2011, 07:41 PM   #7
Newbie
 
Joined: Mar 2011

Posts: 22
Thanks: 0

Re: Extremum search

Ok, I understand it now.

Thank you very much for explaining this
thescratchy is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
extremum, search



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Average search cost of an unsuccessful search in a BST? flexdec Applied Math 0 July 16th, 2013 06:25 AM
Derivative at extremum jkh1919 Calculus 3 July 24th, 2012 07:49 PM
A local extremum problem rubik2008 Calculus 2 July 1st, 2012 06:37 PM
derivative of extremum anigeo Calculus 2 May 12th, 2012 07:50 AM
Extremum StevenMx Calculus 3 December 16th, 2008 11:13 AM





Copyright © 2018 My Math Forum. All rights reserved.