Finding critical points of function of 2 variables

[RM5152]

I have been asked to find the critical points of the function x^3+3xy^2-12x+6y so I found partial derivatives and set to 0 but I don’t know how to solve the function I’m left with (namely x^3-4x-1=0) .. any help would be appreciated. Thanks

 

[Dr.Peterson]

I have been asked to find the critical points of the function x^3+3xy^2-12x+6y so I found partial derivatives and set to 0 but I don’t know how to solve the function I’m left with (namely x^3-4x-1=0) .. any help would be appreciated. Thanks

It will be a little easier if you don't drop the exponent:



(and use the simplified form, too).

It still won't be pretty, but you can do it.

 

[Steven G]

Please divide that equation by 3. It just makes me nervous still seeing it there.

x^2 + (1/x)^2 = 12

(x + 1/x)^2 = 14

x+1/x = +/-sqrt(14)

x^2 -sqrt(14)x + 1= 0 or x^2 +sqrt(14)x + 1= 0

Continue....

 

[RM5152]

Please divide that equation by 3. It just makes me nervous still seeing it there.

x^2 + (1/x)^2 = 12

(x + 1/x)^2 = 14

x+1/x = +/-sqrt(14)

x^2 -sqrt(14)x + 1= 0 or x^2 +sqrt(14)x + 1= 0

Continue....

Does this way work .. if I divide the equation by 3 will I not get x^2+(1/x^2) - 4 ?? I will do it this way if the way I have shown is not correct. Thanks

 

[Dr.Peterson]

Does this way work .. if I divide the equation by 3 will I not get x^2+(1/x^2) - 4 ?? I will do it this way if the way I have shown is not correct. Thanks

Yours is the method I used.

Don't forget to check whether each point satisfies the equations, and classify the critical points.

 

[Dr.Peterson]

Don't forget to check whether each point satisfies the equations, and classify the critical points.

Just because it's too nice not to share, here is a graph showing level curves at z=0 and z=15, and the four critical points:

​

This strongly suggests what you'll find for each point. Here's the 3D graph (with rescaled z):

​

 

[nasi112]

Does this way work .. if I divide the equation by 3 will I not get x^2+(1/x^2) - 4 ?? I will do it this way if the way I have shown is not correct. Thanks

It is nice that you found the four critical points. Also Dr.Peterson's colorful graph suggests that.

FYI, these four critical points represent, two saddle points, a local minima, and a local maxima.

 

[Steven G]

Does this way work .. if I divide the equation by 3 will I not get x^2+(1/x^2) - 4 ?? I will do it this way if the way I have shown is not correct. Thanks

Please divide that equation by 3. It just makes me nervous still seeing it there.

x^2 + (1/x)^2 = 12

(x + 1/x)^2 = 14

x+1/x = +/-sqrt(14)

x^2 -sqrt(14)x + 1= 0 or x^2 +sqrt(14)x + 1= 0

Continue....

Yes, you will get -4 and not -12. I didn't divide the 12 by 3!

 

[topsquark]

Yes, you will get -4 and not -12. I didn't divide the 12 by 3!

Why would you have divide 12 by 3!? That would give you 12/3! = 12/6 = 2.

-Dan