x≥0,y≥0,z≥0,p=-3x+y+2z,q=x-2y+4z,x+y+z=1,则点(p,q)的活动范围为多少

Question

x≥0,y≥0,z≥0,p=-3x+y+2z,q=x-2y+4z,x+y+z=1,则点(p,q)的活动范围为多少

Answer 1

解方程组x+y+z＝1,
-3x+y+2z＝p,
x-2y+4z＝q.
得27x=8-6p+q,
27y=14+3p-5q,
27z=5+3p+4q.
图7-41由x≥0,y≥0,z≥0
得点（p,q）满足的约束条件为：
8-6p+q≥0,
14+3p-5q≥0,
5+3p+4q≥0.
利用线行规划可得结果