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Find the local minima and maxima of function:

$$f(x,y) = x^2-2x+y^2$$

It's easy task with one-variable functions. What should I go about in this case?

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min is $-1$ at $(x,y)=(1;0)$ – Le Chifre Feb 3 '13 at 17:27
up vote 2 down vote accepted

\begin{align} f(x,y)&=x^2 - 2x+y^2\\ \nabla f(x,y)&=\begin{bmatrix} 2x-2\\2y \end{bmatrix}\\ \nabla^2f(x,y)&=\begin{bmatrix} 2&0\\0&2\end{bmatrix}\\ \text{Set : }\nabla f(x,y)&=0\\ \implies (x,y)&=(1,0) \end{align} This is Minima. (Hessian is Positive Definite)

The function has no maxima since Hessian can never be Negative Definite. This is also obvious from the graph of the function.


enter image description here

(From Google Search)

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+10 Nice graph. :-) – Babak S. Feb 3 '13 at 17:26

Hint: First of all find the critical points by doing : $$f_x=0,~f_y=0$$ Assume $(a,b)$ is such that oint. Now find the following terms: $$\Delta_1=f_{xx},~~\Delta_2=f_{xx}f_{yy}-f_{xy}$$ Now if $$\Delta_1|_{(a,b)}>0,~\Delta_2|_{(a,b)}>0$$ then $(a,b)$ will make $f$ minimum. If $$\Delta_1|_{(a,b)}<0,~\Delta_2|_{(a,b)}>0$$ then $(a,b)$ will make $f$ maximum. And when $$\Delta_2|_{(a,b)}<0$$ then $(a,b)$ will make $f$ a saddle point.

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Nicely explained! + 1 (Yay! > 12K) – amWhy Feb 4 '13 at 1:04

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