# Critical Points in 3D

1. Oct 5, 2009

### calorimetry

1. The problem statement, all variables and given/known data
Find the critical points of the function f(x, y) = sinx + siny + cos(x+y)
where 0<=x<=pi/4 and 0<=y<=pi/4

2. Relevant equations
First and second order partial derivative of f(x, y)

3. The attempt at a solution
To find the critical points, I first find the first partial derivative with respect to x and y.

fx(x,y) = cosx - sin(x+y)
fy(x,y) = cosy - sin(x+y)

Set both of the first partial derivative = 0
cosx = sin(x+y) x=pi/4 and y=0
cosy = sin(x+y) x=0 and y=pi/4

Here is where I got stuck, I noticed two points where the above equations are true, but how do I find all the critical points?

I checked the two points I found using
D = fxxfyy - (fxy)^2
fxx = -sinx - cos(x+y)
fyy = -siny - cos(x+y)
fxy = -cos(x+y)

and I found that D for both points are less than zero, which suggest that they both are saddle points. I then graphed the function f(x, y) and these two points don't appear to be saddle points on f. I assume that I made some mistake somewhere either in the derivative or the graph, if someone can check this for me, it would be greatly appreciated.

2. Oct 5, 2009

### LCKurtz

You said:

fx(x,y) = cosx - sin(x+y)
fy(x,y) = cosy - sin(x+y)

Set both of the first partial derivative = 0
cosx = sin(x+y) x=pi/4 and y=0
cosy = sin(x+y) x=0 and y=pi/4

But you must solve these simultaneously. You have picked values that solve them separately, which is irrelevant. But what you can get from those last two equations is that cos(x) = cos(y). And on [0, π/4] that means x = y. Use that in your equation for fx = 0 to see if you can find critical points in your domain.

Depending on what your problem asks for, you may need to check the boundaries too.:yuck:

3. Oct 5, 2009

### calorimetry

Thanks LCKurtz, I haven't solve it yet but your suggestion is really helpful. It makes a lot more sense when I look at the graph. The maximum of the surface is indeed a point where x and y are equal.