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Finding local min, max, and saddle points in multivariable calculus

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woodenbox
#1
Oct24-10, 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 with--does 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.
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Telemachus
#2
Oct24-10, 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.
woodenbox
#3
Oct24-10, 09:55 PM
P: 9
ah, that makes sense. thanks for your help.


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