What are the critical points of #f(x,y) = x^3 + y^3 - xy#?

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Answer 1 · 2015-11-22T04:18:09

#(0,0)# and #(1/3,1/3)#

#f_x = 3x^2-y#
#f_y = 3y^2-x#

Setting both partial derivatives to #0# and solving yields:

#3x^2-y=0# #rArr# #y=3x^2#

Substituting in the other equation, we get:

#3(3x^2)^2-x=0#

So, #27x^4-x=0# which entails that #x=0# or #x=1/3#.

Use #y=3x^2# (or the symmetry of the function) to get the points #(0,0)# and #(1/3,1/3)#