Finding the critical points of a multivariable function

Click For Summary

Homework Help Overview

The discussion revolves around finding the critical points of the multivariable function f(x, y) = x³ + y⁴ - 21x - 18y² and using the Second Derivative Test to classify these points. Participants are exploring the necessary steps to derive the critical points and the implications of their findings.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to solve the partial derivatives for x and y to find critical points, questioning the number of solutions expected. There is discussion about the potential for multiple solutions and the implications of missing solutions.

Discussion Status

Some participants have provided partial solutions and noted errors in previous attempts, suggesting that there are additional solutions to consider. The conversation reflects a collaborative effort to clarify misunderstandings and explore different approaches to the problem.

Contextual Notes

Participants express uncertainty regarding the number of critical points and the validity of their solutions, particularly in relation to the requirement for six different critical point answers. There is also mention of the need to consider solutions that may have been overlooked.

Koi9
Messages
11
Reaction score
0

Homework Statement

\
Find the critical points of the function. Then use the Second Derivative Test to classify each critical point (or state that the test fails).

f(x, y) = x3 + y4 - 21x - 18y2




The Attempt at a Solution



partial x derivative=3x^2-21=0

partial y= 4y^3-36y=0

i try to solve those for numbers, but the answer space in the question has space for 6 different critical point answers, so I am not sure where to go from here.
 
Physics news on Phys.org
Koi9 said:

Homework Statement

\
Find the critical points of the function. Then use the Second Derivative Test to classify each critical point (or state that the test fails).

f(x, y) = x3 + y4 - 21x - 18y2




The Attempt at a Solution



partial x derivative=3x^2-21=0

partial y= 4y^3-36y=0

i try to solve those for numbers, but the answer space in the question has space for 6 different critical point answers, so I am not sure where to go from here.
What did you get when you solved these for numbers?

You should get two solutions for x and three solutions for y.
 
SammyS said:
What did you get when you solved these for numbers?

You should get two solutions for x and three solutions for y.

partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2

must be missing something, because I don't really see another way to solve it.Thanks,
Matt
 
Last edited:
Assignment due soon, anyone help please?
 
Koi9 said:
partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2

must be missing something, because I don't really see another way to solve it.


Thanks,
Matt

That should be x=\pm\sqrt{\frac{21}{3}}=\pm\sqrt{7}\,.

Similarly, use ± for the y solution.

Also, when you divided by y, you got rid of the y=0 solution.
 
Koi9 said:
partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2
In addition to what SammyS said, there's another error here.
Setting fy = 0 yields
4y3-36y=0
==> 4y(y2 - 9) = 0
I hope you can see that this equation has three solutions, none of which is 3/2.
Koi9 said:
must be missing something, because I don't really see another way to solve it.


Thanks,
Matt
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 20 ·
Replies
20
Views
1K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K