Find all the critical points for this function?

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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).