# Finding max and min of a function of several variables

1. Oct 18, 2009

### tnutty

Finding max and min of a function

1. The problem statement, all variables and given/known data

Find min , max , and saddle points of the function :

f(x,y) = e^x*cos(y)

f_x = e^x*cos(y)
f_xx = e^x*cos(y)
f_xy = -e^x * sin(y)

f_y = -e^x * sin(y)
f_y = -e^x * cos(y)

First finding the critical points :

f_x = e^x*cos(y)
f_y = -e^x * sin(y)

I know that e^x will never be 0 , so either cos(y) or sin(y) has to be 0;

cos(y) = 0 , when y = (2n+1)*pi/2
sin(y) = 0 , when y = n*pi

where n is a natural number.

how would I proceed next ?

Last edited: Oct 18, 2009
2. Oct 18, 2009

### Dick

At this point I would start considering the possibility that f(x,y) doesn't have any critical points.

3. Oct 18, 2009

### tnutty

Yea you are right, in the back of the book it says NONE, but how do I justify that?

4. Oct 18, 2009

### Dick

You want cos(y)=0 AND sin(y)=0 to have a critical point. Look at the conditions you've figured out for that possibility. Can they ever BOTH be true?

5. Oct 18, 2009

### tnutty

cos(y) = 0 , when y = (2n+1)*pi/2
sin(y) = 0 , when y = n*pi

(2n+1)*pi/2 = n*pi

pi*n + pi/2 = n*pi
pi*n - pi*n = -pi/2

0 = -pi/2

so it seems like there is no solution or no critical point.

Thanks a lot Dick for your help. I really mean it.

6. Oct 18, 2009

### Dick

Very welcome. But you've got a technical flaw. You shouldn't assume both n's are the same. You want to look for solutions of (2m+1)*pi/2 = n*pi where n and m might be different integers. That's (m-n)pi+pi/2=0. Still no solution, of course. Just a heads up.

7. Oct 18, 2009

### tnutty

Amazing, smart people bring smile to my face.