Finding local max, min and saddle points

  • Thread starter Thread starter maff is tuff
  • Start date Start date
  • Tags Tags
    Local Max Points
Click For Summary
SUMMARY

The discussion focuses on finding local maxima, minima, and saddle points for the function f(x,y) = (1 + xy)(x + y). The user successfully derived the first and second partial derivatives: f_{x} = 1 + 2xy + y^2, f_{y} = 1 + 2xy + x^2, f_{xx} = 2y, f_{yy} = 2x, and f_{xy} = 2(x + y). The critical points were determined by setting the first partial derivatives to zero, leading to the realization that the symmetry in the equations allows for the conclusion that y can equal -x, which ultimately helped in finding the correct critical points.

PREREQUISITES
  • Understanding of multivariable calculus, specifically partial derivatives
  • Familiarity with critical point analysis in functions of two variables
  • Knowledge of symmetry in mathematical functions
  • Ability to solve polynomial equations
NEXT STEPS
  • Study the method for finding critical points in multivariable functions
  • Learn about the second derivative test for classifying critical points
  • Explore symmetry in functions and its implications for optimization
  • Practice solving polynomial equations involving multiple variables
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and optimization techniques, as well as anyone involved in mathematical problem-solving related to multivariable functions.

maff is tuff
Messages
65
Reaction score
1

Homework Statement



f(x,y)=(1+xy)(x+y)



Homework Equations





The Attempt at a Solution



I started out by expanding and got:

x+y+x^2y+xy^2

Then I found all my partial derivatives and second derivatives:

f_{x}=1+2xy+y^2, f_{y}=1+2xy+x^2, f_{xx}=2y, f_{yy}=2x, f_{xy}=2(x+y)

I know that both first partial derivatives must equal zero so I get:

f_{x}=1+2xy+y^2=0 and f_{y}=1+2xy+x^2=0

This is the part I am stuck at; I can't find the critical points. I notice that there is symmetry so I tried subtracting the equations but I got y=x and got:

f_{x}=1+2x(x)+(x)^2=1+2x^2+x^2=0=1+3x^2=0---->x^2=-\frac{1}{3}

I also tried setting f_{x} and f_{y} equal to each other but that didn't seem to work.

Thanks in advance for the help
 
Last edited by a moderator:
Physics news on Phys.org
I think I may have got it. When I got that x^2=y^2 I didn't account for that y could equal (-x). I did that and it got the right answer.
 

Similar threads

Replies
1
Views
2K
Replies
5
Views
1K
Replies
9
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
Replies
20
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K