My Math Forum Extremum search

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 March 5th, 2011, 02:51 PM #1 Newbie   Joined: Mar 2011 Posts: 22 Thanks: 0 Extremum search Hello, I have the following exercise: $f(x,y)= x^2 - 2xy + 2y^2$ I want to search the extremum with help of a matrix. $\frac{d}{dx} = 2x - 4y \frac{d}{dy} = -2x + 4y$ So I get the following matrix: $\begin{array}{cc} \frac{d^2f}{d^2x} & \frac{d^2f}{dxdy}\\ \frac{d^2f}{dydx} & \frac{d^2f}{d^2y}\\ \end{array} = \begin{array}{cc} 2 & -2 \\ -2 & 4\\ \end{array}$ 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: $\begin{array}{cc} 2 & 0 \\ 0 & 2 \\ \end{array}$ 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) = x²-2xy+2y² Check your work. $\frac{\partial f}{\partial x}=2x-2y,\quad\frac{\partial f}{\partial y}=-2x+4y$ 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 $\frac{\partial^2f}{\partial x^2}$ 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 $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$. Second partial derivative test If the determinant is positive, then the critical point is a maximum (if $\frac{\partial^2 f}{\partial x^2}<0$) or a minimum (if $\frac{\partial^2 f}{\partial x^2}>0$). 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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