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I am trying to find the local min/max and saddle points of the function: $f(x,y) = 9 - 2x + 4y - x^{2} - 4y^{2}$

This is what I have so far:

$f_{x}(x,y) = -2 - 2x = -2 (x + 1) \Rightarrow x + 1 = 0 \Rightarrow x = -1$

$f_{y}(x,y) = 4 - 8y = -4 (2y - 1) \Rightarrow 2y - 1 = 0 \Rightarrow y = 0.5$

Any ideas on how to proceed?

  • 2
    By noticing (completing the square) that $f(x,y) = 9 - 2x + 4y - x^2 - 4y^2= 11-(x+1)^2-4(y-\frac12)^2$ you can see immediately that $f(x,y)\le 11$ and $f(x,y)=11$ only for $x=-1$, $y=\frac12$. Hence, it is local maximum; the function looks like a paraboloid. However, for more complicated function, using second derivatives could be better. http://en.wikipedia.org/wiki/Second_partial_derivative_test2011-03-29

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