Multiple Variable Min Max Question

Click For Summary
SUMMARY

The discussion centers on finding the critical points of the function z=(x^5)y+(xy^5)+xy. The partial derivatives are calculated as fx(x,y)=(5x^4)y+(y^5)+y and fy(x,y)=x^5+(5xy^4)+x. After factoring, the critical point is determined to be solely at (x,y)=(0,0), as the second factors in both equations cannot equal zero due to their even powers. This conclusion is reached through the analysis of the equations and their properties.

PREREQUISITES
  • Understanding of multivariable calculus, specifically critical point analysis.
  • Familiarity with partial derivatives and their applications.
  • Knowledge of factoring polynomials in two variables.
  • Basic concepts of even and odd functions.
NEXT STEPS
  • Study the method of Lagrange multipliers for constrained optimization.
  • Learn about the second derivative test for classifying critical points.
  • Explore the implications of even and odd functions in calculus.
  • Investigate the use of software tools like Mathematica for symbolic computation of critical points.
USEFUL FOR

Students and educators in multivariable calculus, mathematicians focusing on optimization problems, and anyone interested in understanding critical point analysis in mathematical functions.

stau40
Messages
37
Reaction score
0

Homework Statement


Find the critical point for z=(x^5)y+(xy^5)+xy


Homework Equations


fx(x,y)=(5x^4)y+(y^5)+y=0
fy(x,y)=x^5+(5xy^4)+x=0

The Attempt at a Solution


After finding fx and fy shown above, I attempt to find the critical points in one of the equations above, but the only number that works (that I can think of) is x=0 or y=0 and this doesn't seem correct to me after substituting it into the other equation. Am I doing something wrong?
 
Physics news on Phys.org
You aren't doing anything wrong. Factor a y out of the first equation and an x out of the second. Can you make an argument that x=y=0 is the only critical point?
 
After factoring I'm left with:

fx(x,y)=y(5x^4+y^4+1)=0
fy(x,y)=x(x^4+5y^4+1)=0

Since both equations are to an even power (can't be negative), it brings me to the conclusion that only x=y=0 works.
 
stau40 said:
After factoring I'm left with:

fx(x,y)=y(5x^4+y^4+1)=0
fy(x,y)=x(x^4+5y^4+1)=0

Since both equations are to an even power (can't be negative), it brings me to the conclusion that only x=y=0 works.

Right. The second factors can never be zero. So the first must.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
6K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
6
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K