Finding the critical points of a multivariable function

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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.
 

Answers and Replies

  • #2
SammyS
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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.
 
  • #3
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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:
  • #4
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Assignment due soon, anyone help please?
 
  • #5
SammyS
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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 [itex]x=\pm\sqrt{\frac{21}{3}}=\pm\sqrt{7}\,.[/itex]

Similarly, use ± for the y solution.

Also, when you divided by y, you got rid of the y=0 solution.
 
  • #6
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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.
must be missing something, because I don't really see another way to solve it.


Thanks,
Matt
 

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