Calculating critical points and classifying them

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SUMMARY

The discussion focuses on finding and classifying critical points for the function f(x,y) = sin(x)sin(y)sin(x+y) within the bounds 0 ≤ x, y ≤ π. The critical points are determined using the equations x = π/3(2n - m) and y = π/3(2m - n), where n ≥ 1 and m ≤ 2, both belonging to the integer set (Z). The classification of these points is performed using the second derivative test, specifically evaluating the determinant fxxfyy - fxy² to identify saddle points and other classifications.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly critical points.
  • Familiarity with the second derivative test for classifying critical points.
  • Knowledge of trigonometric functions and their properties.
  • Ability to work with integer sets and inequalities in mathematical expressions.
NEXT STEPS
  • Study the second derivative test in detail to understand its applications and limitations.
  • Explore the properties of trigonometric functions in multivariable calculus.
  • Learn about alternative methods for classifying critical points, such as the Hessian matrix.
  • Investigate the implications of critical points in optimization problems.
USEFUL FOR

Students and educators in mathematics, particularly those studying multivariable calculus, as well as anyone interested in optimization techniques and critical point analysis.

Gekko
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Homework Statement



Find all critical points and classify them


Homework Equations



f(x,y) = sin(x)sin(y)sin(x+y)

0<=x,y<=Pi


The Attempt at a Solution



fx=sinysin(2x+y) and fy=sinxsin(2y+x)

Therefore critical points are at:

x=Pi/3(2n-m) , y=Pi/3(2m-n) where n>=1, m<=2, n,m belong to integer set (Z)

fxx = sin(2x+2y)-sin(2x)
fyy = sin(2x+2y)-sin(2y)
fxy = sin(2x+2y)

Now, to classify the critical points I was simply going to test for:

fxxfyy-fxy^2

If <0 it is a saddle point etc and substitute for x the definitions above to obtain the equation we can use to classify with

Is this correct? Is there a better way? Would greatly appreciate comments. Thanks
 
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Yep, that's the second derivative test. So it's definitely correct (assuming you have that "etc" part right in your comment). As for a better way...there might be something intuitive about this function in particular that makes it special. I don't see anything, but I'm not sure. However, in general, the second derivative test is the way to go.
 

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