Posted by Lucy on Tuesday, February 28, 2012 at 4:51pm.
F(x,y)=-3x^2y-3xy^2+36xy
For maximum/minimum,Fx=0 AND Fy=0
Fx=36y-6xy-3y^2=3y(12-2x-y)=0 ...(1)
Fy=36x-6xy-3x^2=3x(12-2y-x)=0 ...(2)
Clearly (0,0) is a critical point.
If y=0, solve for x in (2) to get x=12, so
(12,0) is another critical point.
Similarly (0,12) is another critical point.
Finally, solve for x,y in
12-2x-y=0 and
12-x-2y=0 to get
x=y=4, or (4,4) is a critical point.
So the four critical points are
(0,0)
(0,12)
(12,0)
(4,4)
Second derivative test
Calculate
Fxx=-6y
Fyy=-6x
Fxy=Fyx=6(6-x-y) [Clairot's theorme]
Now calculate
D(x,y)=Fxx.Fyy-Fxy²
=36xy-6(36-12(x+y)+(x+y)^2)
D(0,0)=-216<0 => saddle point
D(0,12)=-216<0 => saddle point
D(12,0)=-216<0 => saddle point
D(4,4)=552 >0 => max or min
Fxx=-6(4)=-24 <0 => maximum