How do I find the possible y values in the following equation?

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The discussion centers on solving the equation 32y^3 - 168y + 100 = 0 to find potential y values. The user derived this equation while attempting to find critical points of the surface defined by the function f(x,y) = 10yx^2 - 5x^2 - 4y^2 - x^4 - 2y^4. The partial derivatives ∂f/∂x and ∂f/∂y were calculated, leading to the substitution that resulted in the cubic equation. Participants suggest that while analytical solutions exist, a numerical approach may suffice for finding y values.

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How do I find the potential y values for 32y^3-168y+100=0

It's a multi-variable calc problem. I have to find the critical points of the surface
f(x,y) = 10yx^2 - 5x^2 - 4y^2-x^4-2y^4

∂f/∂x = 20xy-10x-4x^3 = 0
x=0 or x^2 = (20y-10)/4

∂f/∂y = 8y^3+8y-10x^2=0

I subbed x^2 in and got 32y^3-168y+100=0

Any ideas what I'm doing wrong if I am? Or do I have to find the y values that satisfy 32y^3-168y+100=0?

Thanks for reading, guys.

Edit: Just realized I should've posted this in the homework help section. Don't know how to delete. Mod could you move this for me?
 
Last edited:
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Pizzerer said:
How do I find the potential y values for 32y^3-168y+100=0

It's a multi-variable calc problem. I have to find the critical points of the surface
f(x,y) = 10yx^2 - 5x^2 - 4y^2-x^4-2y^4

∂f/∂x = 20xy-10x-4x^3 = 0
x=0 or x^2 = (20y-10)/4

∂f/∂y = 8y^3+8y-10x^2=0

I subbed x^2 in and got 32y^3-168y+100=0

Any ideas what I'm doing wrong if I am? Or do I have to find the y values that satisfy 32y^3-168y+100=0?

Thanks for reading, guys.

Edit: Just realized I should've posted this in the homework help section. Don't know how to delete. Mod could you move this for me?
You can report your own post to get it moved.

While there are formulas for those equations, I doubt that you have to use them. A numerical solution could be fine. The case x=0 is easier to study.
 

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