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).
