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Finding local min, max, and saddle points in multivariable calculus 
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#1
Oct2410, 09:40 PM

P: 9

1. The problem statement, all variables and given/known data
Find the local maximum and minimum values and saddle point(s) of the function. f(x,y) = (e^x) * cosy The textbook answer is "none." I sort of reached that conclusion but I don't know if my reasoning is correct... 2. Relevant equations The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b)  [fxy(a,b)]^2 if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum if D < 0 then f(a,b) is a saddle point if D = 0 then the test is inconclusive 3. The attempt at a solution Since the textbook answer is none, I suppose in my answer I should demonstrate how I came to the conclusion that there are no mins, maxes, or saddle points. I worked something out that seems to lead to that conclusion, but I'm not sure if my reasoning makes sense. That's the thing I really would like help withdoes this line of reasoning lead to the correct conclusion? So... I've took the first partial derivatives and set them equal to zero: fx = (e^x) * cosy = 0 fy = (e^x) * siny = 0 and since e^x never reaches 0, the only points where these statements can be true is where cosy = 0 and/or siny = 0 which occurs at pi/2, pi/2 and 0 Next, take the relevant second partials: fxx = (e^x)*cosy fyy = (e^x)*cosy fxy = (e^x)*siny D = (e^x)cosy*(e^x)cosy  [(e^x)siny]^2 = (e^2x)*cos(y)^2  (e^x)*sin(y)^2 = (e^2x) * [ cos(y)^2 + sin(y)^2 ] = (e^2x) Since D is always < 0 (since e^2x is always positive), this analysis suggests there is a continuous line of saddle points, which is not possible/doesn't make sense. So, there must not be any local mins, maxes, or saddle points. Is it correct to reason that way? Any help, advice, or criticism would be much appreciated. 


#2
Oct2410, 09:47 PM

P: 533

I think that here:
fx = (e^x) * cosy = 0 fy = (e^x) * siny = 0 Is the key. Because those partial derivatives are never zero simultaneously. Which means that there are no critical points. 


#3
Oct2410, 09:55 PM

P: 9

ah, that makes sense. thanks for your help.



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