Register to reply

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

Share this thread:
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?

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.
Phys.Org News Partner Science news on
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
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.
Oct24-10, 09:55 PM
P: 9
ah, that makes sense. thanks for your help.

Register to reply

Related Discussions
Finding local min, max, and saddle points in multivariable calculus Calculus & Beyond Homework 3
Local min/max/saddle points of 3d graphs Calculus & Beyond Homework 1
Multivariable Calculus: finding relative extrema/saddle points Calculus & Beyond Homework 2
Finding the critical points of a multivariable function and determining local extrema Calculus & Beyond Homework 1
Finding Critical Points and Local Extrema of a Multivariable Function Calculus & Beyond Homework 4