Trig Function Critical points

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SigepBrandon

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I'm sure you guys are getting tired of seeing posts about this, I apologize, I'm just having a hard time with it.. I appreciate the help.

The problem reads:
Find the critical points of f(x,y)= sin(x) + sin(y) + cos(x+y) for \(\displaystyle 0\leq x \leq\frac{\pi }{4}\) and \(\displaystyle 0\leq y \leq\frac{\pi }{4}\). Classify each as a local max, local min, or a saddle point.

Well.. I know the first step. Find the partials.
Fx(x,y)= d/dx (sin(x) + sin(y) + cos(x+y)) = cos(x) - sin(x+y)
Fy(x,y)= d/dy (sin(x) + sin(y) + cos(x+y)) = cos(y) - sin(x+y)

I know that the critical points are going to be on the boundary, or at the critical points. The critical points occur when the first derivatives are = 0.

solving for y: cos(x) - sin(x+y) = 0
This is where I get stuck. I messed around a bit with exchanging trig identities, but just wasted a lot of time, because I don't really know what I'm doing or if that is the correct approach.

Also, I set the partials equal to each other hoping that would help.

cos(x) - sin(x+y) = cos(y) - sin(x+y)
adding sin(x+y) to both sides
cos(x)=cos(y) => x=y

I think that tells me that I only have to solve for x or y, but not sure how to do either..
 
SigepBrandon said:
I'm sure you guys are getting tired of seeing posts about this, I apologize, I'm just having a hard time with it.. I appreciate the help.

Brandon, as you can see, I am a JUNIOR member, and I try to avoid commenting on the calculus page because I am incredibly rusty. You have been waiting a while without an answer so I shall take a shot, confident that someone will correct me if I lead you astray.

The problem reads:
Find the critical points of f(x,y)= sin(x) + sin(y) + cos(x+y) for \(\displaystyle 0\leq x \leq\frac{\pi }{4}\) and \(\displaystyle 0\leq y \leq\frac{\pi }{4}\). Classify each as a local max, local min, or a saddle point.

Well.. I know the first step. Find the partials.

Here is where you go astray I think. I at least would get rid of that cos(x + y) before doing anything else. I am not sure that it is necessary, but I am quite sure that it will simplify things. That is, I'd use trig identities before fussing with the calculus.
So, I'D START WITH cos(x + y) = [cos(x) * cos(y)] - [sin(x) * sin(y)].

See if that helps. If not, I'm sure galactus or someone will be around shortly and clean up my mess.

Fx(x,y)= d/dx (sin(x) + sin(y) + cos(x+y)) = cos(x) - sin(x+y)
Fy(x,y)= d/dy (sin(x) + sin(y) + cos(x+y)) = cos(y) - sin(x+y)

I know that the critical points are going to be on the boundary, or at the critical points. The critical points occur when the first derivatives are = 0.

solving for y: cos(x) - sin(x+y) = 0
This is where I get stuck. I messed around a bit with exchanging trig identities, but just wasted a lot of time, because I don't really know what I'm doing or if that is the correct approach.

Also, I set the partials equal to each other hoping that would help.

cos(x) - sin(x+y) = cos(y) - sin(x+y)
adding sin(x+y) to both sides
cos(x)=cos(y) => x=y

I think that tells me that I only have to solve for x or y, but not sure how to do either..
Click to expand...
 
Thanks for your reply Jeff. I think using the identity first complicates the derivative quite a bit. I did use a similar identity when solving the first partial derivatives for 0. but not much farther than that.

the work is as follows,
Solve cos(y)-sin(x+y)=0
=Cos(y)-[sin(x) cos(y) + cos(x) * sin(y)]=0
cos(y)=sin(x)*cos(y)+cos(x)*sin(y)

other than drawing a map of how a bunch of identities fit together, that's about as far as I got with it...
 
but to do it jeff's way:

f(x,y)=sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y))
df/dx sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y)) = cos(x)*(-sin(y))-sin(x)*cos(y)+cos(x) = 0 =>
x can not be 0 because of the last term cos(x)
df/dy sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y)) = (-sin(x))*cos(y)-cos(x)*sin(y)+cos(y) = 0 =>
y can not be 0 because of the last term cos(y)

?
 
conditions are:

cos(x) = cos(y) = 0

and

sin(x) + sin(y) = 1
 
Thanks sub!

For sin(x) + sin(y) = 1; x and y would be pi/6.

For cos(x) = cos(y) = 0; x and y would have to = pi/2 but that is outside my given range for x and y. (they have to be between 0 and pi/4)

Thanks to some software I know that a max occurs at (1/2,1/2) and that would meet the first condition, but can you explain a bit more? I do not see how you arrived there...
 
SigepBrandon said:
but to do it jeff's way:

f(x,y)=sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y))
df/dx sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y)) = cos(x)*(-sin(y))-sin(x)*cos(y)+cos(x) = 0 =>
x can not be 0 because of the last term cos(x)
df/dy sin(x)+sin(y)+(cos(x)*cos(y)-sin(x)*sin(y)) = (-sin(x))*cos(y)-cos(x)*sin(y)+cos(y) = 0 =>
y can not be 0 because of the last term cos(y)

?
Click to expand...

cos(x)*(-sin(y))-sin(x)*cos(y)+cos(x) = 0 ................. and cos(x) = cos(y)

cos(x)*(-sin(y))-sin(x)*cos(x)+cos(x) = 0

cos(x) * [ 1 - sin(x) -sin(y)] = 0

and so on ......
 
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