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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?

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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. – Martin Sleziak Mar 29 '11 at 19:08

$f_y=-2(x+1), f_y=-8(y-1/2)$, so the only critical point is $(-1,1/2)$. now you need to decide whether this is max/min/saddle. you can apply the "second derivative test" but the function is so simple that it is easier to just complete the square wrt both $x$ and $y$ to get $$(11-z)=(x+1)^2+4(y-1/2)^2.$$ this is the equation of an elliptic paraboloid that is opening downward (most calc books have a section on quadratic surfaces, and comparing functions to quadratic polynomials is essentially the "second derivative test"). hence $(-1,1/2)$ is a MAX.

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I think you should get $-(x+1)^2$ on the RHS. – Martin Sleziak Mar 29 '11 at 19:24
The minus sign was shifted to the LHS; there was a numerical error (now corrected) in the constant (also moved to the LHS), but the equation is otherwise the same as that in your comment above. – RecklessReckoner Apr 11 '14 at 3:06

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