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

1 Answer
Jun 30, 2015

Answer:

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))#