LeGrange Multipliers Finding critical points of function

Click For Summary
SUMMARY

The discussion focuses on finding the critical points of the function f(x,y) = x + y² under the constraint g(x,y) = x² + y² = 1 using the method of Lagrange multipliers. The equations derived include ∇f = λ∇g, leading to the relationships f_x = 1 = 2λx and f_y = 2y = 2λy. The critical points are determined by solving for x and y, resulting in two scenarios: when y = 0, x can be derived from the constraint, and when λ = 1, x = 1/(2λ) allows for further calculation of y. The solution process is clarified by confirming that if y = 0, λ is irrelevant, and x can be directly calculated from the constraint.

PREREQUISITES
  • Understanding of Lagrange multipliers
  • Familiarity with gradient vectors (∇f and ∇g)
  • Knowledge of solving equations involving constraints
  • Basic calculus, particularly partial derivatives
NEXT STEPS
  • Study the method of Lagrange multipliers in depth
  • Practice solving optimization problems with multiple constraints
  • Explore applications of critical points in real-world scenarios
  • Learn about the geometric interpretation of Lagrange multipliers
USEFUL FOR

Students in calculus or optimization courses, mathematicians, and anyone interested in applying Lagrange multipliers to find critical points of functions under constraints.

Saladsamurai
Messages
3,009
Reaction score
7

Homework Statement



Find the critical points of [itex]f(x,y)=x+y^2[/itex] subject to the constraint [itex]g(x)=x^2+y^2=1[/itex]



Homework Equations


[itex]\nabla f=\lambda\nabla g[/itex]
[itex]g(x,y)=1[/itex]



The Attempt at a Solution



[tex]f_x=1=2\lambda*x\Rightarrow x=\frac{1}{2\lambda}[/tex]

[tex]f_y=2y=2\lambda*y\Rightarrow y(\lambda-1)=0\Rightarrow y=0 \\ or\\ \lambda=1[/tex]

[tex]x^2+y^2=1[/tex]

I am a little confused as to where I go from here?
 
Physics news on Phys.org
Saladsamurai said:

Homework Statement



Find the critical points of [itex]f(x,y)=x+y^2[/itex] subject to the constraint [itex]g(x)=x^2+y^2=1[/itex]



Homework Equations


[itex]\nabla f=\lambda\nabla g[/itex]
[itex]g(x,y)=1[/itex]



The Attempt at a Solution



[tex]f_x=1=2\lambda*x\Rightarrow x=\frac{1}{2\lambda}[/tex]

[tex]f_y=2y=2\lambda*y\Rightarrow y(\lambda-1)=0\Rightarrow y=0 \\ or\\ \lambda=1[/tex]

[tex]x^2+y^2=1[/tex]

I am a little confused as to where I go from here?

You've done the hard part. If y= 0, then [itex]\lambda[/itex] doesn't matter: Find x from x2+ y2= 1. If [itex]\lambda= 1[/itex], then x= 1/2 and you can find y from x2+ y2= 1.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K