已知x>3,y>0,且2x+xy-3y=60, 求2x+3y之最小值?

已知x>3，y>0，,且2x+xy－-3y=60，求2x+3y之最小值?
Sol
2x+xy－-3y=60
(2x+xy)－-3y=60
x(y+2)－-3(y+2)=54
(x－3)(y+2)=54
x－3>0，y+2>0
(2x+3y)/2
=[2(x－3)+3(y+2)]/2
>=√[6*(x－3)(y+2)]
=√(6*54)
=18
2x+3y>=36

x=12,
y=4,
2x+3y=36,

2x+xy-3y=60
(x-3)(y+2)=54 ---> 以 x=3, y+2=0 為漸進線之雙曲線
(x-3)(y+2)=54=(3*3)(3*2)
在 x-3=9 ---> x=12,
y+2=6 ---> y=4 時, 2x+3y有最小值 24+12=36