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？

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？

先画图，在取点

取点A（1,3）代入得Zmax=3x+2y=3+6=9

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

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