Finding minimum value of function with two variable

Click For Summary

Discussion Overview

The discussion revolves around finding the minimum value of a cost function that depends on two variables, x and y. Participants explore methods for determining the optimal values of these variables through calculus, particularly using partial derivatives and critical points.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant describes a cost calculation formula involving two variables and seeks to find the minimum cost values for x and y.
  • Another participant requests the specific function and work done so far to provide more targeted assistance.
  • There is a discussion about setting the first partial derivatives to zero to find critical points and whether these points represent minimum or maximum values.
  • Participants discuss the second derivative test and the Hessian matrix as methods to determine the nature of the critical points.
  • One participant expresses a need to confirm that substituting the found values of x and y into the original cost equation will yield the minimum cost.
  • There is a request for help in determining the range of x and y values that would yield a solution, referencing a figure that is not clearly understood by others.
  • Participants express confusion regarding the figure and suggest that the question about the figure may be a separate issue from the original problem.

Areas of Agreement / Disagreement

Participants generally agree on the methods for finding critical points and testing for minima or maxima, but there is no consensus on the clarity of the figure presented or its relevance to the original cost calculation problem.

Contextual Notes

The discussion includes limitations related to the clarity of the visual aids provided and the definitions of variables used in the context of the problem.

gevni
Messages
25
Reaction score
0
I have a formula for cost calculation that contain x and y two variable. I have to find the value of (x,y) where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
I took 1st partial derivative with respect to x and then with y and found the value of x and y where function will give minimum value. My question is how to combine both to get one answer as optimal (x,y) where function has minimum cost?
 
Last edited:
Physics news on Phys.org
Please type out your function and your work so far.
 
gevni said:
I have a formula for cost calculation that contain x and y two variable. I have to find the value of (x,y) where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
I took 1st partial derivative with respect to x and then with y and found the value of x and y where function will give minimum value. My question is how to combine both to get one answer as optimal (x,y) where function has minimum cost?
I assume you set the partial derivatives to 0 and solved the two resulting equations for x and y. Now what do mean by "combine both to get one answer? The one point, (x, y) IS the point where f(x, y) is minimum (or maximum- you need to check which). For example if $f(x,y)= (x-3)^2+(y- 4)^2$ then setting the partial derivatives equal to 0 gives $2(x- 3)= 0$ and $2(y- 4)= 0$ so x= 3 and y= 4. The "critical point" is (3, 4) where the value of the function is f(3, 4)= 0, a minimum.
 
Country Boy said:
I assume you set the partial derivatives to 0 and solved the two resulting equations for x and y. Now what do mean by "combine both to get one answer? The one point, (x, y) IS the point where f(x, y) is minimum (or maximum- you need to check which). For example if $f(x,y)= (x-3)^2+(y- 4)^2$ then setting the partial derivatives equal to 0 gives $2(x- 3)= 0$ and $2(y- 4)= 0$ so x= 3 and y= 4. The "critical point" is (3, 4) where the value of the function is f(3, 4)= 0, a minimum.
Thank you! yes after getting partial derivatives with x and y I set the resulting equations to 0 for getting values of x and y and got the formula for x= ...and y=...
Just want to confirm that now I have to replace x and y in original cost equation with these formulas (x= and y=) and it will results in minimum cost for that cost equation right?.
One more question how do I know that these values of x and y are for minimum?
 
One way is to compare the value at that point to values close by. Another is to use the "second derivative test": if $(f_{xx})(f_{yy})-(f_{xy})^2$ is positive at the given point, then it is either a maximum or a minimum. If it is negative then the point is a "saddle point".

(This is the determinant of the "Hessian matrix", $\begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}$. It is actually the product of the two eigenvalues. Each eigenvalue tells you if the graph curves up (if the eigenvalue is positive) or down (if negative) in the x or y directions. The determinant is positive if they have the same sign, both positive or both negative, negative if they have different signs.)
 
Last edited:
Country Boy said:
One way is to compare the value at that point to values close by. Another is to use the "second derivative test": if $(f_{xx})(f_{yy})-(f_{xy})^2$ is positive at the given point, then it is either a maximum or a minimum. If it is negative then the point is a "saddle point".

Thank you I compared points closed by and get to know the minimum. Now there is one more thing, I have to provide the range of x and y within which if their values fall then there is a solution else there will be no solution. Can you please look into the attached figure it is n*n square divided into A,B and C area, can you please help me finding the upper range for both variables (x,y)?

0<x< ?
0<y<?
 

Attachments

  • 110461264_1177808379226535_2654335419636091325_n.jpg
    110461264_1177808379226535_2654335419636091325_n.jpg
    38.5 KB · Views: 112
That picture is very hard to understand. You have region labeled "A", "B", "C", and "A" and are apparently using those letters to represent the areas of those regions but don't say that. In addition you have an "h" (or is it "n"?) but do not define it. Is h the length of a side of the square? Finally, what in the world does this picture have to do with the original problem, a "cost calculation"?
 
Country Boy said:
That picture is very hard to understand.
I think I batter had to ask it as a separate question (with detail) as it is a different problem. Many thanks for your help :)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K