Let \ \ h(x,y)=x^3-6x^2-3y^2\ \ . Find all critical points of \ \ f\ \ ? Then classify each as either a relative maximum, minimum, saddle point, or neither.

1 Answer
Narad T.
Jun 3, 2018

Please see the explanation below

Explanation:

![www.slideshare.net](useruploads.socratic.org)

The function is

h(x,y)=x^3-6x^2-3y^2

The first derivatives are

(delh)/(delx)=3x^2-12x-0

(delh)/(delx)=0, =>, 3x^2-12x=3x(x-4)=0

<=>, {(x=0),(x=4):}

(delh)/(dely)=-6y

(delh)/(dely)=0, =>, y=0

The second derivatives are

(del^2h)/(delx^2)=6x^2-12

(del^2h)/(dely^2)=-6

(del^2h)/(delxdely)=0

(del^2h)/(delydelx)=0

The Hessian of h(x,y) is

D^2h(x,y)=|((del^2h)/(delx^2),(del^2h)/(delydelx)),((del^2h)/(delxdely),(del^2h)/(dely^2))|

=|(6x^2-12,0),(0,-6)|

=-6(6x^2-12)

Therefore,

D^2h(0,0)=72 and (del^2h(0,0))/(delx^2)=-12

The point (0,0) corresponds to a relative maximum.

D^2h(4,0)=-504 and (del^2h(4,0))/(delx^2)=84

The point (4,0) corresponds to a saddle pint.

Related questions
Impact of this question
2051 views around the world
You can reuse this answer
Creative Commons License