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

1 Answer
Jun 30, 2015

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))