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

Jun 30, 2015

Set both partial first derivatives to $0$ and solve the system.

#### Explanation:

$f \left(x , y\right) = 2 {y}^{3} + 3 {x}^{3} - 6 x y$

Partial derivatives set to $0$

${f}_{x} \left(x , y\right) = 9 {x}^{2} - 6 y = 0$

$f \left(x , y\right) = 6 {y}^{2} - 6 x = 0$

Solve the system

From the first equation, we get: $y = \frac{3}{2} {x}^{2}$
Substituting into the second equation gives us:

$6 {\left(\frac{3}{2} {x}^{2}\right)}^{2} - 6 x = 0$

So $6 \left(\frac{9}{4} {x}^{4}\right) - 6 x = 0$

And $\frac{27}{2} {x}^{4} - 6 x = 0$

Clear the fraction: $27 {x}^{4} - 12 x = 0$

So we get: $3 x \left(9 {x}^{3} - 4\right) = 0$

Whose solutions are: $x = 0$ and $x = \sqrt[3]{\frac{4}{9}}$

Recall that: $y = \frac{3}{2} {x}^{2}$, so

when $x = 0$, $y = 0$. Thus $\left(0 , 0\right)$ is a critical point.

when $x = \sqrt[3]{\frac{4}{9}}$,
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 $\left(\sqrt[3]{\frac{4}{9}} , \sqrt[3]{\frac{2}{3}}\right)$ is a critical point.

The critical points are $\left(0 , 0\right)$ and $\left(\sqrt[3]{\frac{4}{9}} , \sqrt[3]{\frac{2}{3}}\right)$