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

1 Answer
Apr 29, 2017

Answer:

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

Explanation:

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.