Question

How do you find critical points for `f(x,y)= 2y^3+3x^3-6xy`?

Answer 1

Set both partial first derivatives to `0` and solve the system.

Explanation:

`f(x,y)= 2y^3+3x^3-6xy`

Partial derivatives set to `0`

`f_x(x,y)= 9x^2-6y = 0`

`f(x,y)= 6y^2-6x = 0`

Solve the system

From the first equation, we get: `y = 3/2x^2`
Substituting into the second equation gives us:

`6(3/2x^2)^2 - 6x =0`

So `6(9/4x^4) - 6x =0`

And `27/2x^4-6x=0`

Clear the fraction: `27x^4-12x=0`

So we get: `3x(9x^3-4)=0`

Whose solutions are: `x=0` and `x=root(3)(4/9)`

Recall that: `y = 3/2x^2`, so

when `x=0`, `y=0`. Thus `(0,0)` is a critical point.

when `x=root(3)(4/9)`,
`y=3/2root(3)((4/9)^2) =3/2 root(3)[(2^2*2)/(3^3*3)] = 3/2*2/3root(3)(2/3) = root(3)(2/3`.
Thus `(root(3)(4/9), root(3)(2/3))` is a critical point.

Answer
The critical points are `(0,0)` and `(root(3)(4/9), root(3)(2/3))`