For maximum/minimum,Fx=0 AND Fy=0
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-x-2y=0 to get
x=y=4, or (4,4) is a critical point.
So the four critical points are
Second derivative test
Fxy=Fyx=6(6-x-y) [Clairot's theorme]
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
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