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

Nov 22, 2015

$\left(0 , 0\right)$ and $\left(\frac{1}{3} , \frac{1}{3}\right)$

#### Explanation:

${f}_{x} = 3 {x}^{2} - y$
${f}_{y} = 3 {y}^{2} - x$

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

$3 {x}^{2} - y = 0$ $\Rightarrow$ $y = 3 {x}^{2}$

Substituting in the other equation, we get:

$3 {\left(3 {x}^{2}\right)}^{2} - x = 0$

So, $27 {x}^{4} - x = 0$ which entails that $x = 0$ or $x = \frac{1}{3}$.

Use $y = 3 {x}^{2}$ (or the symmetry of the function) to get the points $\left(0 , 0\right)$ and $\left(\frac{1}{3} , \frac{1}{3}\right)$