Question

设x,y满足条件x-y+2≥0,2x+y-5≤0,y≥2,则z=3x+2y的最大值为？

Answer 1

先画图，在取点

取点A（1,3）代入得Zmax=3x+2y=3+6=9

Answer 2

x-y+2≥0 (1')
2x+y-5≤0 (2')
y≥2 (3')

x-y+2=0 (1)
2x+y-5=0 (2)
y=2 (3)

case 1: (1) and (2)
x-y+2=0 (1)
2x+y-5=0 (2)

(1)+(2)
x=1

from (1)
1-y+2=0
y= 3

y≥2 (3')
(x,y)=(1,3) 满足 (3')
z=3x+2y =3(1)+2(3) =9

case 2: (1) and (3)
x-y+2=0 (1)
y=2 (3)

2x+y-5≤0 (2')
(x,y)=(0,2) 满足 (2')

z=3x+2y =3(0)+2(2) =5

case 3: (2) and (3)
2x+y-5=0 (2)
y=2 (3)

x-y+2≥0 (1')
(x,y)=(3/2,2) 满足 (1')

z=3x+2y =3(3/2)+2(2) =9/2 +4 = 17/2

max z=3x+2y = case1 =9

Answer 3

最大值为9

Answer 4