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

Question

9259 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-22T04:18:09

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

$f_{x} = 3 x^{2} - y$ $f_x = 3x^2-y$
$f_{y} = 3 y^{2} - x$ $f_y = 3y^2-x$

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

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

Substituting in the other equation, we get:

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

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

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

What are the critical points of f(x,y) = x^3 + y^3 - xyf(x,y)=x3+y3−xyf(x,y) = x^3 + y^3 - xy?