# How to find the critical points of a function f(x,y)=xy^2-3x^2-y^2+2x+2?

Jun 11, 2015

The critical points are $\left(x , y\right) = \left(1 , - 2\right) , \left(x , y\right) = \left(1 , 2\right)$, and $\left(x , y\right) = \left(\frac{1}{3} , 0\right)$.

#### Explanation:

The partial derivatives of $z = f \left(x , y\right) = x {y}^{2} - 3 {x}^{2} - {y}^{2} + 2 x + 2$ are $\setminus \frac{\setminus \partial z}{\setminus \partial x} = {y}^{2} - 6 x + 2$ and $\setminus \frac{\setminus \partial z}{\setminus \partial y} = 2 x y - 2 y = 2 y \left(x - 1\right)$.

Setting these equal to zero gives a system of equations that must be solved to find the critical points: ${y}^{2} - 6 x + 2 = 0 , 2 y \left(x - 1\right) = 0$.

The second equation will be true if $y = 0$, which will lead to the first equation becoming $- 6 x + 2 = 0$ so that $6 x = 2$ and $x = \frac{1}{3}$, making one critical point $\left(x , y\right) = \left(\frac{1}{3} , 0\right)$.

The second equation of the system above will also be true if $x = 1$, which will lead to the first equation becoming ${y}^{2} - 4 = 0$ and ${y}^{2} = 4$, making $y = \setminus \pm 2$ and leading to two critical points $\left(x , y\right) = \left(1 , 2\right) , \left(x , y\right) = \left(1 , - 2\right)$.

You didn't ask for this, but we can also classify these critical points as follows:

1) Find the second-order partials: $\setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial {x}^{2}} = - 6 , \setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial {y}^{2}} = 2 x - 2$, and $\setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial x \setminus \partial y} = \setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial y \setminus \partial x} = 2 y$.

2) Find the discriminant $D = \setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial {x}^{2}} \cdot \setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial {y}^{2}} - {\left(\setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial x \setminus \partial y}\right)}^{2} = 12 - 12 x - 4 {y}^{2}$

3) Plug the critical points into the discriminant: $D \left(1 , 2\right) = 12 - 12 - 16 = - 16$, $D \left(1 , - 2\right) = 12 - 12 - 16 = - 16$, and $D \left(\frac{1}{3} , 0\right) = 12 - 4 - 0 = 8$.

4) Since $D \left(1 , \setminus \pm 2\right) = - 16 < 0$, the critical points at $\left(1 , \setminus \pm 2\right)$ are saddle points.

5) Since $D \left(\frac{1}{3} , 0\right) = 8 > 0$ and $\setminus \frac{\setminus {\partial}^{2} z}{\setminus \partial {x}^{2}} {|}_{\left(\frac{1}{3} , 0\right)} = - 6 < 0$, the critical point at $\left(\frac{1}{3} , 0\right)$ is a local maximum.

Here's a contour map of this function in the $x y$-plane along with its critical points.