Finding Critical Points Of A Function

In summary, WolframAlpha agrees with Mark44 that the error affects your second derivatives. You need to recompute the partial derivatives and determine if the point is a min or a max.
  • #1
Lancelot59
646
1
I'm trying to find the minimia and maxima of the following function without using LaGrange multipliers:

[tex]f(x,y)=sin(x)+sin(y)+sin(x+y)[/tex]
where:
[tex]0\leq x \leq 2\Pi[/tex]
[tex]0\leq y \leq 2\Pi[/tex]

Partial derivatives:
[tex]f_{x}=cos(y)+cos(x+y)[/tex]
[tex]f_{y}=cos(x)+cos(x+y)[/tex]

[tex]f_{xx}=-sin(x+y)[/tex]
[tex]f_{xy}=-sin(y)-sin(x+y)[/tex]
[tex]f_{yy}=-sin(x+y)[/tex]

Now I have no clue how to get all the critical points. I simplified it using fx=0, fy=0, to get fx=fy. Equating and simplifying I got cos(x)=cos(y), x=y.

Now this is where I get lost. How do you pick points to try and solve the equations? I could say that cos(x)=cos(y), then cos(x)-cos(y)=0. So you could pick x or y to be pi/2 or 3pi/2.

However the solution manual to my textbook does not use any of these points...so how does this work?
 
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  • #2
Lancelot59 said:
I'm trying to find the minimia and maxima of the following function without using LaGrange multipliers:

[tex]f(x,y)=sin(x)+sin(y)+sin(x+y)[/tex]
where:
[tex]0\leq x \leq 2\Pi[/tex]
[tex]0\leq y \leq 2\Pi[/tex]

Partial derivatives:
[tex]f_{x}=cos(y)+cos(x+y)[/tex]
[tex]f_{y}=cos(x)+cos(x+y)[/tex]
You got off to a bad start. The two first partials are incorrect.
fx = cos(x) + cos(x + y)
fy = cos(y) + cos(x + y)

Because of this error, there are errors in your second partials as well.
Lancelot59 said:
[tex]f_{xx}=-sin(x+y)[/tex]
[tex]f_{xy}=-sin(y)-sin(x+y)[/tex]
[tex]f_{yy}=-sin(x+y)[/tex]

Now I have no clue how to get all the critical points. I simplified it using fx=0, fy=0, to get fx=fy. Equating and simplifying I got cos(x)=cos(y), x=y.

Now this is where I get lost. How do you pick points to try and solve the equations? I could say that cos(x)=cos(y), then cos(x)-cos(y)=0. So you could pick x or y to be pi/2 or 3pi/2.

However the solution manual to my textbook does not use any of these points...so how does this work?
 
  • #3
Mark44 said:
You got off to a bad start. The two first partials are incorrect.
fx = cos(x) + cos(x + y)
fy = cos(y) + cos(x + y)

Because of this error, there are errors in your second partials as well.
WolframAlpha agrees with me.
Partial X
Partial Y
 
  • #4
Lancelot59 said:
WolframAlpha agrees with me.
Partial X
Partial Y

Take a careful look at what you wrote for the first derivatives in your first post, what Mark44 wrote, and what wolframalpha is giving you. I assure you wolframalpha agrees with Mark44 (and that the error affects your second derivatives).
 
  • #5
DAH! That would do it...

So if fx=fy

cos(x)+cos(x+y)=cos(y)+cos(x+y)
cos(x)=cos(y)
x=y

Now what?
 
  • #6
Lancelot59 said:
DAH! That would do it...

So if fx=fy

cos(x)+cos(x+y)=cos(y)+cos(x+y)
cos(x)=cos(y)
x=y

Now what?

Plug y=x back into f_x=0 to figure out what x (and hence y) must be. To determine if the point is a min or a max you need to recompute the partial derivatives, as the second derivatives in the first post are incorrect because you initially mixed up f_x and f_y. Then, calculate the value of the Hessian matrix to see if it's a min, max or saddle.
 
  • #7
I got it. Thanks!
 

FAQ: Finding Critical Points Of A Function

1. What is the definition of a critical point?

A critical point of a function is a point on the graph where the derivative is equal to zero or undefined. It can also be defined as a point where the slope of the tangent line is zero.

2. Why are critical points important in calculus?

Critical points are important because they help us determine the maximum and minimum values of a function. They also help us identify points where the function changes from increasing to decreasing or vice versa.

3. Can a function have more than one critical point?

Yes, a function can have multiple critical points. This occurs when the derivative is equal to zero or undefined at more than one point on the graph.

4. How do you find critical points of a function?

To find critical points, we need to take the derivative of the function and set it equal to zero. Then, we solve for the values of x that make the derivative equal to zero or undefined. These values will be the critical points of the function.

5. Can critical points be located on the endpoints of a closed interval?

Yes, critical points can be located on the endpoints of a closed interval. It is important to check the endpoints when finding critical points because they can also be maximum or minimum values of the function.

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