Find all the critical points for this function?

enter image source here

1 Answer
Mar 9, 2018

#(0,-2)# is a saddle point
#(-5,3)# is a local minimum

Explanation:

We are given #g(x,y)=3x^2+6xy+2y^3+12x-24y#

First, we need to find the points where #(delg)/(delx)# and #(delg)/(dely)# both equal 0.

#(delg)/(delx)=6x+6y+12#
#(delg)/(dely)=6x+6y^2-24#

#6(x+y+2)=0#
#6(x+y^2-4)=0#

#x+y+2=0#
#x=-y-2#

#-y-2+y^2-4=0#

#y^2-y-6=0#

#(y-3)(y+2)=0#

#y=3 or -2#

#x=-3-2=-5#
#x=2-2=0#

Critical points occur at #(0,-2)# and #(-5,3)#

Now for classifying:
The determinant of #f(x,y)# is given by #D(x,y)=(del^2g)/(delx^2)(del^2g)/(dely^2)-((del^2g)/(delxy))^2#

#(del^2g)/(delx^2)=del/(delx)((delg)/(delx))=del/(delx)(6x+6y+12)=6#

#(del^2g)/(dely^2)=del/(dely)((delg)/(dely))=del/(dely)(6x+6y^2-24)=12y#

#(del^2g)/(delxy)=del/(delx)((delg)/(dely))=del/(delx)(6x+6y^2-24)=6#

#(del^2g)/(delyx)=del/(dely)((delg)/(delx))=del/(dely)(6x+6y+12)=6#

#D(x,y)=6(12y)-36#

#D(0,-2)=72(-2)-36=-180#
#D(-5,3)=72(3)-36=180#

Since #D(0,-2)<0#, #(0,-2)# is a saddle point.
And since #D(-5,3)>0 and (del^2g)/(delx^2)>0#, #(-5,3)# is a local minimum. (#(del^2g)/(delx^2)=6# so we don't need to do any calculations).