帮我做一道数学题？ 30
2x+4y-2x-y<=20-11解出y<=3
4x+2y-x-2y<=22-10解出x<=4
因为y的利润最高，所有y的值要尽可能大
所以y取3，x取4
2x+y≤11 (1')
x+2y≤10 (2')
x≥0 (3')
y≥0 (4')
max z=8x+10y
solution
2x+y=11 (1)
x+2y=10 (2)
x=0 (3)
y=0 (4)
case 1: (1) and (2)
2x+y=11 (1)
x+2y=10 (2)
2(2) -(1)
3y = 9
y=3
from (1)
x=4
(x,y)=(4,3) 满足 (3'), (4')
z=8x+10y
z(4,3) = 32+30 =62
case 2: (1) and (3)
2x+y=11 (1)
x=0 (3)
(x,y) = (0, 11)
(x,y)=(0,11) 不满足 (2')
case 2 舍去
case 3: (1) and (4)
2x+y=11 (1)
y=0 (4)
(x,y)=(11/2, 0)
(x,y)=(11/2, 0) 满足 (2'), (3')
z=8x+10y
z(11/2, 0) = 44+0 =44
case 3: (2) and (3)
x+2y=10 (2)
x=0 (3)
(x,y) = (0,5)
(x,y)=(0, 5) 满足 (1'), (4')
z=8x+10y
z(0, 5) =0+50 =50
case 4: (2) and (4)
x+2y=10 (2)
y=0 (4)
(x,y) = (10,0)
(x,y)=(10, 0) 不满足 (1')
case 4 ： 舍去
case 5 ：(3) and (4)
(x,y)=(0,0)
(x,y)=(0, 0) 满足 (1'), (2')
z=8x+10y
z(0, 0) =0+0 =0
所以最大Z ： case （1）
z(4,3) = 32+30 =62
