Question

已知x>0,y>0,且2x+3y-4xy=0,求3x+2y的最小值

Answer 1

用拉哥朗日乘子法：设乘子为t,
F(x,y,t)=3*x+2*y+t(2*x+3*y-4*x*y)
对x,y,t分别求导并等于0，F'x=3+2t-4ty=0;F'y=2+3t-4tx=0,F't=2*x+3*y-4*x*y=0,三式联立解得：x=5/4,y=5/4,t=1,另解舍
则3x+2y=（3+2）*5/4=25/4，这就是最小值

Answer 2

2x+3y-4xy=0
(4y-2)x=3y
x>0，y>0，因此4y-2>0 y>1/2
x=3y/(4y-2)
3x+2y=9y/(4y-2) +2y
=(8y²+5y)/(4y-2)
=[8y²-4y+9y -(9/2) +(9/2)]/(4y-2)
=2y+ 9/4 +(9/2)/(4y-2)
=(2y-1) +(9/4)[1/(2y-1)] +13/4
y>1/2 2y-1>0，由均值不等式得
(2y-1)+(9/4)[1/(2y-1)]≥2√(9/4)=3
3x+2y≥3+ 13/4 =25/4