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

Question

2817 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-29T13:00:20

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

Evaluate the partial derivatives of the first order:

$(del f)/dx =2x+4$

$(del f)/dy =2y$

so the critical points are the solutions of the equations:

${(2x+4=0),(2y=0):}$

${(x=-2),(y=0):}$

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

$(del^2 f)/(dx^2) = 2$

$(del^2 f)/(dx dy) = 0$

$(del^2 f)/(dy^2) = 2$

$abs H = 2 xx 2 -0 = 4$

so we have that:

$abs H > 0$

$[(del^2 f)/(dx^2)]_((x,y) = (-2,0)) >0$

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

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