Solving a min max problem in 2 variables using Mathematica

Click For Summary

Homework Help Overview

The discussion revolves around a min-max problem in two variables, specifically analyzing the behavior of a function defined in R² without a specified domain. Participants are exploring the existence of maximum and minimum points based on the function's derivatives.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the implications of the function's derivatives being equal to zero and questioning the definitions of maxima and saddle points in the context of a two-variable function. There is also a suggestion to consider a coordinate transformation to simplify the analysis.

Discussion Status

The discussion is active, with participants questioning the original poster's reasoning and exploring different interpretations of the function's behavior. Some guidance has been offered regarding the nature of critical points and the relationship between the function's graphical representation and its mathematical properties.

Contextual Notes

There is a lack of consensus on the existence of maximum and minimum points, with some participants suggesting that the function does not have any, while others are probing the implications of critical points where the derivatives are zero.

Wi_N
Messages
119
Reaction score
8
Homework Statement
Solving min max problem in 2 variables using Mathematica
Relevant Equations
(x + y)*Sin[x - y],

3 questions: how many min/max points in R^2 and how many "saddle points"(in 3d)?
Drawing the graph in 3d you see endless "mountains and valleys" which logic tells me there will also be infinite max min points in 2d regardless of where you slice the graph. apparently this is wrong and there is a finite max/min points in R^2/2D. Please note this problem does not have a domain.

f'x= (x + y) Cos[x - y] + Sin[x - y]==0
f'y =-(x + y) Cos[x - y] + Sin[x - y]==0

fx==fy
y =(-x)

f'x=sin2x==0 (infinite max min )

f'y=sin-2y==0 (infinite max min)

...what am i doing wrong?
 
Physics news on Phys.org
Wi_N said:
Drawing the graph in 3d
Like this ? :
1579019321001.png

Wi_N said:
apparently this is wrong
Where do you see the maxima ? (how are maxima defined ?)
Wi_N said:
"saddle points"(in 3d)?
Any definition of such a thing ? For a function of two variables ? Or do you mean 'for z = (x + y)*Sin[x - y] ' ?

What do you call 2D and what 3D ?

Note: You could also make a little coordinate transformation: u = x+y, v = x-y and study u * sin v
 
Wi_N said:
fx==fy
y =(-x)

f'x=sin2x==0 (infinite max min )
f'y=sin-2y==0 (infinite max min)

...what am i doing wrong?
In the first place: What are you doing at all ?

Poor man's math:
maxima is where the rain accumulates when you hold the surface upside down

Mathematical criterion has to do with derivatives allright, but not in the way you treat it.
 
Ya I'm pretty sure I've done it correctly in terms of finding min and max points of a function with two variables.
Take the derivative with respect to each variable independently. and both ==0 you find the critical point. you can use various techniques to find out if its max or min.

Edit: Apparently the answer is 0, the function doesn't have any max or min points...
 
Last edited:
Wi_N said:
function doesn't have any max or min points
That becomes clear from the picture. So what is it with the points that have ##f_x=f_y=0## ?
 
Zero in, zero in! :oldwink:

You got x = -y.

Isn't that or is that a solution to part of the problem? If not why not?

And anyway doesn't it reduce the problem to one in one variable only? :oldwink:

You can eliminate between your two equations for zero derivatives in more than one way. :oldwink:
 
Last edited:
Wi_N said:
Edit: Apparently the answer is 0, the function doesn't have any max or min points...
Yes. Can you see that from the picture ?
And what about the saddle points ?
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
6K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K