Critical points and 2nd partials test

Click For Summary
SUMMARY

The discussion focuses on finding critical points and testing for relative extrema of the function f(x,y) = 12xy - x²y - 2xy² using the 2nd partials test. The critical point identified is (5,3). Initial calculations for the second partial derivatives yield fxx = -2, fyy = -4, and fxy = 0, leading to a determinant d = 0, which initially suggested no conclusion. However, a correction reveals that d = 8, confirming a local minimum at the critical point (5,3).

PREREQUISITES
  • Understanding of multivariable calculus concepts, specifically critical points.
  • Familiarity with the 2nd partial derivatives test for classifying extrema.
  • Knowledge of how to compute first and second partial derivatives of functions.
  • Ability to solve equations involving multiple variables.
NEXT STEPS
  • Study the method for finding critical points in multivariable functions.
  • Learn about the implications of the determinant in the 2nd partials test.
  • Explore examples of functions with multiple critical points and their classifications.
  • Review the concepts of local minima and maxima in the context of multivariable calculus.
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable functions, as well as anyone seeking to understand the application of the 2nd partials test in determining relative extrema.

camino
Messages
42
Reaction score
0

Homework Statement



Find all critical points of f(x,y) = 12xy-x^2 y-2xy^2 and test them for relative extrema using the 2nd partials test.

Homework Equations





The Attempt at a Solution



{0=fx(x,y)=12-2x-2 10x-2 x=5
{0=fy(x,y)=12-4y y=3

Critical point: (5,3)

fxx(x,y)= -2
fyy(x,y)= -4
fxy(x,y)= 0

d(5,3)=fxx(5,3)fyy(5,3)-(fxy(5,3))^2
d=(0)(0)-(0)
d=0 no conclusion
---------------------------------------

Just wondering if I did this right?
 
Physics news on Phys.org
camino said:

Homework Statement



Find all critical points of f(x,y) = 12xy-x^2 y-2xy^2 and test them for relative extrema using the 2nd partials test.

Homework Equations





The Attempt at a Solution



{0=fx(x,y)=12-2x-2 10x-2 x=5
{0=fy(x,y)=12-4y y=3

Critical point: (5,3)

fxx(x,y)= -2
fyy(x,y)= -4
fxy(x,y)= 0

d(5,3)=fxx(5,3)fyy(5,3)-(fxy(5,3))^2
d=(0)(0)-(0)
d=0 no conclusion
---------------------------------------

Just wondering if I did this right?
fxx(5,3)fyy(5,3)= (-2)(-4), not (0)(0)!
 
Ok, so

d=(-2)(-4)-(0)
d=8

Is that right?
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
7K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K