# Finding the critical points of a multivariable function

## 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.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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.

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

Mark44
Mentor
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