# How do you find critical points for function of two variables f(x,y)=8x^3+144xy+8y^3?

Mar 27, 2015

For two-variables function, critical points are defined as the points in which the gradient equals zero, just like you had a critical point for the single-variable function $f \left(x\right)$ if the derivative $f ' \left(x\right) = 0$. The matter is that you now can differentiate the function with respect to more than one variable (namely 2, in your case), and so you must define a derivative for each directions.

The gradient is thus defined as the $n$-dimensional vector (again, in your case $n = 2$), and its coordinates are the derivatives with respect to each variable. So, the gradient of a two-variable function $f \left(x , y\right)$ is the vector
$\left(\setminus \frac{\setminus \partial f}{\setminus \partial x} , \setminus \frac{\setminus \partial f}{\setminus \partial y}\right)$, where deriving with respect to a variable means to consider the other as a constant.

Let's compute the two derivatives:
$\setminus \frac{\setminus \partial f}{\setminus \partial x} = 24 {x}^{2} + 144 y$
$\setminus \frac{\setminus \partial f}{\setminus \partial y} = 144 x + 24 {y}^{2}$

To find the critical points, we must find the values of $x$ and $y$ for which
$\left(\setminus \frac{\setminus \partial f}{\setminus \partial x} , \setminus \frac{\setminus \partial f}{\setminus \partial y}\right) = \left(0 , 0\right)$ holds.
In other words, we must solve
$24 {x}^{2} + 144 y = 0$
$24 {y}^{2} + 144 x = 0$
Simplifying both expression, we have
${x}^{2} + 6 y = 0$
${y}^{2} + 6 x = 0$
An obvious solution is $\left(x , y\right) = \left(0 , 0\right)$, which is thus our first critical point, while the other solution are $\left(x , y\right) = \left(- 6 , - 6\right)$, which is the second and last critical point.