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 
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? 
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) = x²2xy+2y² Check your work. but the matrix is correct and the partial derivatives exist everywhere. For a critical point, you need to solve 2x2y=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) = 84 = 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. 
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? 
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?

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. 
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 

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