# What are the values and types of the critical points, if any, of f(x, y) = x^3+y^3-3*x*y-7?

Nov 12, 2017

There is a saddle point at $\left(0 , 0\right)$ and a relative minimum at $\left(1 , 1\right)$

#### Explanation: Here, we have

$f \left(x , y\right) = {x}^{3} + {y}^{3} - 3 x y - 7$

Compute the partial derivatives

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

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

${f}_{x} = 0$, $\implies$, ${x}^{2} = y$

${f}_{y} = 0$, $\implies$, ${y}^{2} = x$

${x}^{4} = {y}^{2} = x$

${x}^{4} - x = 0$

x(x^3-1))=0

$x \left(x - 1\right) \left({x}^{2} + x + 1\right) = 0$

$x = 0$ and $x = 1$

$y = 0$ and $y = 1$

The critical point are $\left(0 , 0\right)$ and $\left(1 , 1\right)$

Now, compute the second partial derivatives

${f}_{x x} = 6 x$

${f}_{y y} = 6 y$

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

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

Therefore,

$D \left(x , y\right) = {f}_{x x} {f}_{y y} - {f}_{x y}^{2} = \left(6 x\right) \left(6 y\right) - {\left(- 3\right)}^{2} = 36 x y - 9$

$D \left(0 , 0\right) = - 9$

This point $\left(0 , 0\right)$ corresponds to a saddle point

D(1,1)=36-9)=27 and ${f}_{x x} \left(1 , 1\right) = 6$

There is a relative min at $\left(1 , 1\right)$