Find all relative maxima/minima and saddle points

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SUMMARY

The discussion focuses on finding critical points, relative maxima, minima, and saddle points for the function f(x,y) = 2x³ + xy² + 5x² + y². The partial derivatives are calculated as fx(x,y) = 6x² + y² + 25x and fy(x,y) = 2xy + 2y. The critical points are determined by setting these derivatives to zero, leading to the equations 2y(x + 1) = 0, which yields y = 0 or x = -1. The user is advised to verify their arithmetic in the calculations of the partial derivatives.

PREREQUISITES
  • Understanding of multivariable calculus, specifically partial derivatives
  • Familiarity with critical point analysis in functions of two variables
  • Knowledge of the second derivative test for classifying critical points
  • Proficiency in algebraic manipulation and solving equations
NEXT STEPS
  • Learn how to apply the second derivative test for functions of two variables
  • Study the method for finding critical points in multivariable functions
  • Explore the implications of saddle points in optimization problems
  • Review examples of similar functions to reinforce understanding of critical point analysis
USEFUL FOR

Students studying multivariable calculus, mathematicians interested in optimization techniques, and educators teaching critical point analysis in functions of two variables.

stolencookie

Homework Statement


f(x,y)=2x^3+xy^2+5x^2+y^2

Homework Equations

The Attempt at a Solution


fx(x,y)=6x^2+y^2+25x
fy(x,y)=2xy+2y
fxx(x,y)=12x+25
fyy(x,y)=2y+2
fxy(x,y)=2y
I found the partial derivatives from the equation. I am stuck at finding the critical points from the first two fx(x,y) and fy(x,y). I equaled 0=6x^2+y^2+25x and solved for y^2. I then got y^2=-6x^2-25x and tried substituting it in, fy(x,y)= 2x(6-x^2-25x)^2+2(-6x^2-25x)^2. which did not work out well as I was able to do in similar problems.
 
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stolencookie said:

Homework Statement


f(x,y)=2x^3+xy^2+5x^2+y^2

Homework Equations

The Attempt at a Solution


fx(x,y)=6x^2+y^2+25x
fy(x,y)=2xy+2y
fxx(x,y)=12x+25
fyy(x,y)=2y+2
fxy(x,y)=2y
I found the partial derivatives from the equation. I am stuck at finding the critical points from the first two fx(x,y) and fy(x,y). I equaled 0=6x^2+y^2+25x and solved for y^2. I then got y^2=-6x^2-25x and tried substituting it in, fy(x,y)= 2x(6-x^2-25x)^2+2(-6x^2-25x)^2. which did not work out well as I was able to do in similar problems.
You have ##f_y = 2xy+2y##. Set that equal to ##0##, and factor it getting ##2y(x+1)=0##. That tells you there are two possibilities, either ##y=0## or ##x=-1##. Go from there.
Edit: Added, check your arithmetic for ##f_x## too.
 
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LCKurtz said:
You have ##f_y = 2xy+2y##. Set that equal to ##0##, and factor it getting ##2y(x+1)=0##. That tells you there are two possibilities, either ##y=0## or ##x=-1##. Go from there.
Edit: Added, check your arithmetic for ##f_x## too.

that was a mistype on the fx either than that thank you.
 

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