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Suppose $f(x,y) = x^2+xy+y^2$. How do you write this as two functions $f_{1}(x,y)$ and $f_{2}(x,y)$? I am trying to use Newton's method for $f(x,y)$.

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Apparently you misunderstood how to compute the gradient. Do you know how to do partial derivatives? –  J. M. Apr 19 '12 at 7:12
    
@J.M.: I know how to do that. But you just need to set the gradients equal to 0? Not the actual function? –  Robb Apr 19 '12 at 7:23
    
Depends on what you're trying to do. Are you optimizing $f(x,y)$? –  J. M. Apr 19 '12 at 7:24
    
@J.M. Yeah trying to find the minimum of $f(x,y)$. –  Robb Apr 19 '12 at 7:31
    
Apply Newton's method to the gradient of $f$, ie, $(\frac{\partial f(x,y)}{\partial x}, \frac{\partial f(x,y)}{\partial y})$. –  copper.hat Apr 19 '12 at 7:38

1 Answer 1

To find the minimum of $f(x,y)$ you set both partial derivatives equal to 0. You got: $$\frac{\partial f}{\partial x}=2x+y=0 \ and \ \frac{\partial f}{\partial y}=2y+x=0$$ The only solution to this system is the set $(x_o,y_o)=(0,0)$. To prove that this point is a total minimum, you calculate the determine of the matrix: $$\left(\begin{array}{cc}\frac{\partial^2 f}{\partial x^2}(x_o,y_o)&\frac{\partial^2 f}{\partial x \partial y}(x_o,y_o)\\\frac{\partial^2 f}{\partial x \partial y}(x_o,y_o)&\frac{\partial^2 f}{\partial y^2}(x_o,y_o)\end{array}\right)=\left(\begin{array}{cc}2&1\\1&2\end{array}\right)$$ This determinant is equal to 3. Due to the fact that $\frac{\partial^2 f}{\partial x^2}(x_o,y_o)=2>0$ and $det=3>0$, the previous matrix is positive definite and therefore the point $(x_o,y_o)$ is a local minimum with value $f(x_0,y_0)=0$.

You can find some theory behind multivariable extrema here: Multivariable extrema

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