# Multivariable Critical Points Question

Why does $$f(x,y) = x^4 + y^4-16xy$$ have critical points $$(0, 0), (2, 2), (-2, -2)$$?

I get the first critical point and was able to solve for $$x$$ all the way until $$x^9/4^4$$ but I do not know what to do after that. Can anyone clarify?

The necessary condition $$\nabla f = 0$$ leads to the equations $$x^3 = 4 y$$ and $$y^3 = 4x$$. Inserting $$y = x^3 /4$$ into the second equation gives $$(x^3/4)^3 = 4 x$$, i.e.,

$$x^9 - 4^4 x = x(x^8 - 4^4 ) = 0.$$

The first factor becomes zero for $$x = 0$$ and the second factor becomes $$0$$ for

$$x^8 = 4^4 = (2^2)^4 = 2^8,$$

which implies $$x = \pm 2$$. For $$x = 2$$, one gets $$y^3 = 8$$, which implies $$y = 2$$. For $$x = -2$$, one similarly gets $$y = -2$$.

EDIT. A shorter way would be to notice that the symmetry of the equations $$x^3 = 4y$$ and $$y^3 = 4x$$ immediately delivers $$x = y$$. Therefore, $$x^3 - 4x = x (x^2 - 4) = 0$$, which implies $$x = 0$$ or $$x = \pm 2$$.

• You can reduce the degree of the resulting equations considerably by taking advantage of the symmetry of the two original equations. – amd Jan 18 at 22:42
• Thank you for the help Jan! – OGK Jan 19 at 8:07