Finding Critical Points for f(x, y) = sinx + siny + cos(x+y)

In summary, the critical points of the function f(x, y) = sinx + siny + cos(x+y) within the given domain of 0<=x<=pi/4 and 0<=y<=pi/4 can be found by solving the simultaneous equations cos(x) = cos(y) and cosx = sin(x+y), which leads to x = y and x = pi/4, y = 0. These points can be confirmed as saddle points by calculating the determinant D and graphing the function. The boundaries may also need to be checked for critical points.
  • #1
calorimetry
30
0

Homework Statement


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


Homework Equations


First and second order partial derivative of f(x, y)


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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 

Related to Finding Critical Points for f(x, y) = sinx + siny + cos(x+y)

1. What are critical points in 3D?

Critical points in 3D refer to points where the first-order partial derivatives of a function are equal to zero. These points can be maxima, minima, or saddle points.

2. How do you find critical points in 3D?

To find critical points in 3D, you must first find the first-order partial derivatives of the function with respect to each variable. Then, set each derivative equal to zero and solve the resulting system of equations to find the critical points.

3. Why are critical points important in 3D?

Critical points are important in 3D because they can help determine the maxima, minima, and saddle points of a function. These points can provide valuable information about the behavior of a function and can be used to optimize a system or solve optimization problems.

4. How are critical points related to the gradient of a function?

The gradient of a function at a point is defined as a vector pointing in the direction of steepest ascent. At critical points, the gradient of a function is equal to zero, indicating that there is no direction of steepest ascent or descent.

5. Can a function have multiple critical points in 3D?

Yes, a function can have multiple critical points in 3D. These points can be local maxima, local minima, or saddle points. It is important to analyze the behavior of the function at each critical point to understand the overall behavior of the function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
620
  • Calculus and Beyond Homework Help
Replies
3
Views
638
  • Calculus and Beyond Homework Help
Replies
8
Views
556
  • Calculus and Beyond Homework Help
Replies
2
Views
562
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
600
  • Calculus and Beyond Homework Help
Replies
1
Views
506
  • Calculus and Beyond Homework Help
Replies
6
Views
988
Back
Top