Finding critical points of a function of 2 variables

[davyjones]

This is originally an optimization/calculus question but I'm stuck on the algebra.

The question is to find all critical points of the function
f(x,y) = x^3 + y^3 - 6*x*y

I found partial derivatives for x and y and set them to zero (i.e. to find flat points on the function surface):

Fx = 3*x^2 - 6*y = 0
Fy = 3*y^2 - 6*x = 0

By setting the equations equal to eachother and factorizing + simplifying I've ended up with:

x^2 + 2*x -y^2 -2*y = 0

My question is how do I take this factorize this further to find the values for x and y that make this true? If you can do this please show your working; I'd like to understand how it was done for future reference. Thanks!

 

[soroban]

Hello, davyjones!

Find all critical points of the function: .\(\displaystyle f(x,y) \:=\: x^3 + y^3 - 6xy\)

I found partial derivatives for x and y and set them to zero.
. . \(\displaystyle \begin{array}{ccccccc}f_x &=& 3x^2 - 6y &=& 0 & [1] \\ f_y &=& 3y^2 - 6x &=& 0 & [2] \end{array}\) . Good!

I'd solve it like this . . .

Solve [1] for \(\displaystyle y\!:\;\;y \:=\:\frac{x^2}{2}\)

Substitute into [2]: .\(\displaystyle 3\left(\frac{x^2}{2}\right)^2 - 6x \:=\:0 \quad\Rightarrow\quad \frac{3}{4}x^4 - 6x \:=\:0\)

. . . \(\displaystyle \frac{3}{4}x(x^3 - 8) \:=\:0 \quad\Rightarrow\quad x \:=\:0,\,2 \quad\Rightarrow\quad y \:=\:0,\,2\)


Critical points: .\(\displaystyle (0,\,0),\;(2,\,2)\)

(To classify the critical points, use the Second Partials Test.)

 

[davyjones]

Thanks a lot soroban, that all makes sense.