Finding Local/Absolute Extrema of f(x,y)=x^2+y^2

  • Thread starter Thread starter chunkyman343
  • Start date Start date
  • Tags Tags
    Extrema
Click For Summary

Homework Help Overview

The discussion revolves around finding the local and absolute extrema of the function f(x,y) = x^2 + y^2, constrained by the triangle defined by the lines x=0, y=0, and y+2x=2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the identification of critical points and the necessity to evaluate the function on the boundary of the defined domain. There is a focus on the implications of finding extrema at critical points versus boundary points.

Discussion Status

Some participants have provided insights into the nature of critical points and the importance of checking boundaries for extrema. There is an ongoing exploration of the relationship between critical points and boundary evaluations, with no clear consensus on the final outcomes yet.

Contextual Notes

Participants are working within the constraints of the specified triangular domain and are considering the implications of their findings on local and absolute extrema.

chunkyman343
Messages
4
Reaction score
0
]I have to find the local/absolute extrema for the following function:
f(x,y)=x^2+y^2

bounded by the triangle x=0,y=0,y+2x=2

So far i have:
fx(x,y)=2x
fy(x,y)=2y
fxx(x,y)=2
fyy(x,y)=2
fxy&fyx(x,y)=0

critical pts at (0,0,0)
domain: 0<=x<=1, 0<=y<=2

i don't know what i should do next?
 
Physics news on Phys.org
Since it's only a function of two dimensions, the critical point you get is actually at (0,0). Notice that there are precisely two points where a maximum or a minimum can occur: At a critical point, or on the boundary of the domain. So now you have to check the boundary for any maxima (that you found a critical point on the boundary is pure coincidence; the rest of the boundary still needs to be checked)
 
So this is what i get:

local max: (0,2,4)
abs. min:(1,0,1)
abs. max:(1,2,5)
 
What happened to the critical point you found earlier?
 
(0,0,0) is an absolute min.

how does that look now?
 
with (1,0,1) being a local min.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K