# How do you find the critical points for f(x, y) = x^2 + 4x + y^2 and the local max and min?

Apr 29, 2017

The point $\left(- 2 , 0\right)$ is a minimum for the function.

#### Explanation:

Evaluate the partial derivatives of the first order:

$\frac{\partial f}{\mathrm{dx}} = 2 x + 4$

$\frac{\partial f}{\mathrm{dy}} = 2 y$

so the critical points are the solutions of the equations:

$\left\{\begin{matrix}2 x + 4 = 0 \\ 2 y = 0\end{matrix}\right.$

$\left\{\begin{matrix}x = - 2 \\ y = 0\end{matrix}\right.$

We have therefore a single critical point: $\left(- 2 , 0\right)$. To determine the character of the point we have to look at the Hessian matrix:

$\frac{{\partial}^{2} f}{{\mathrm{dx}}^{2}} = 2$

$\frac{{\partial}^{2} f}{\mathrm{dx} \mathrm{dy}} = 0$

$\frac{{\partial}^{2} f}{{\mathrm{dy}}^{2}} = 2$

$\left\mid H \right\mid = 2 \times 2 - 0 = 4$

so we have that:

$\left\mid H \right\mid > 0$

${\left[\frac{{\partial}^{2} f}{{\mathrm{dx}}^{2}}\right]}_{\left(x , y\right) = \left(- 2 , 0\right)} > 0$

then the point $\left(- 2 , 0\right)$ is a minimum.