Question

求解二元一次方程

9x+4y<=360
4x+5y<=200
3x+10y<=300

求700x+1200y最大值（x,y＞=0）

Answer 1

9x+4y<=360 (1)
4x+5y<=200 (2)
3x+10y<=300 (3)
解：
5*(1)-4*(2)：29x1<=1000

x1<=1000/29 (4)
2.5*(1)-(3)：19.5x2<=600
x2<=400/13 (5)
2*(2)-(3)： 5x3<=100
x3<=20 (6)
(4)分别代入(1)、(2)、(3) ：9*1000/29+4y1<=360
y1<=360/29
4*1000/29+5y2<=200
y2<=360/29
3*1000/29+10y3<=300
y3<=570/29
y1=360/29
(5)分别代入(1)、(2)、(3) ：9*400/13+4y1<=360
y1<=270/13
4*400/13+5y2<=200
y2<=200/13
3*400/13+10y3<=300
y3<=270/13
y2<=200/13
(6)分别代入(1)、(2)、(3) ：9*20+4y1<=360
y1<=45
4*20+5y2<=200
y2<=24
3*20+10y3<=300
y3<=24
y3<=24
即：x1<=1000/29、y1<=360/29；
x2<=400/13、y2<=200/13；
x3<=20、y3<=24；
求700x+1200y（x,y＞=0）最大值
y取最大值，则700x+1200y（x,y＞=0）有最大值
x=20、y=24
700x+1200y=700*20+1200*24=42800

或

解：
5*(1)-4*(2)：29x1<=1000

x1<=1000/29 (4)
2.5*(1)-(3)：19.5x2<=600
x2<=400/13 (5)
2*(2)-(3)： 5x3<=100
x3<=20 (6)
求700x+1200y（x,y＞=0）最大值
y取最大值(x取最小值)，则700x+1200y（x,y＞=0）有最大值
即：x<=20 (7)
(7)分别代入(1)、(2)、(3)：9*20+4y<=360
y1<=45
4*20+5y<=200
y2<=24
3*20+10y<=300
y3<=24
y<=24
x=20、y=24时700x+1200y（x,y＞=0）有最大值
700x+1200y=700*20+1200*24=42800