# How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for f(x, y) = x^2 + 4x + y^2?

Jul 28, 2015

The unique critical point is $\left(x , y\right) = \left(- 2 , 0\right)$ and it is a local minimum (and the local minimum value of the function is $f \left(- 2 , 0\right)$=4-8+0=-4#.

#### Explanation:

The first-order partial derivatives of $z = f \left(x , y\right) = {x}^{2} + 4 x + {y}^{2}$ are $\frac{\partial z}{\partial x} = 2 x + 4$ and $\frac{\partial z}{\partial y} = 2 y$.

Setting both of these equal to zero results in a system of equations whose unique solution is clearly $\left(x , y\right) = \left(- 2 , 0\right)$, so this is the unique critical point of $f$.

The second-order partials are $\frac{{\partial}^{2} z}{\partial {x}^{2}} = 2$, $\frac{{\partial}^{2} z}{\partial {y}^{2}} = 2$, and $\frac{{\partial}^{2} z}{\partial x \partial y} = \frac{{\partial}^{2} z}{\partial y \partial x} = 0$

This makes the discriminant for the (multivariable) Second Derivative Test equal to

$D = \frac{{\partial}^{2} z}{\partial {x}^{2}} \cdot \frac{{\partial}^{2} z}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} z}{\partial x \partial y}\right)}^{2} = 2 \cdot 2 - {0}^{2} = 4 > 0$,

which means the critical point is either a local max or a local min (it's not a saddle point).

Since $\frac{{\partial}^{2} z}{\partial {x}^{2}} = 2 > 0$, the critical point is a local min.

Jul 28, 2015

See Explanation

#### Explanation:

$f \left(x , y\right) = {x}^{2} + 4 x + {y}^{2}$

${f}_{x} \left(x , y\right) = 2 x + 4$
${f}_{y} \left(x , y\right) = 2 y$

The critical point makes both partial derivatives $0$ (simultaneously).

For this function there is one critical point: $\left(- 2 , 0\right)$

To determine whether $f$ has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.)

${f}_{x x} \left(x , y\right) = 2$

${f}_{x y} \left(x , y\right) = 0 \text{ }$ (As usual, this is also ${f}_{y x} \left(x , y\right)$)

${f}_{y y} \left(x , y\right) = 2$

Evaluate the second partials at the critical point (In this case they are all constant, but in general we cannot skip this step.)

At the critical point $\left(- 2 , 0\right)$, we get

$A = {f}_{x x} \left(- 2 , 0\right) = 2$

$B = {f}_{x y} \left(- 2 , 0\right) = 0$

$C = {f}_{y y} \left(- 2 , 0\right) = 2$

Calculate $D = A C - {B}^{2}$

$D = \left(2\right) \left(2\right) - {\left(0\right)}^{2} = 4$

Apply the second derivative test:

Since $D$ is positive, we look at $A$ and with $D > 0$ and $A > 0$, we have a local minimum at the critical point.

$f \left(- 2 , 0\right) = 4 - 8 = - 4$

To conclude:

$f$ has a local minimum of $- 4$ at $\left(- 2 , 0\right)$.