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设三种等级的酒用于三种商标酒的量分别为x1,x2,x3; y1,y2,y3;z1,z2,z3,e为一个正的小数,可以很小。
Max w=5.5(x1+x2+x3)+5(y1+y2+y3)+4.8(z1+z2+z3)-6(x1+y1+z1)-4.5(x2+y2+z2)-3(x3+y3+z3)
s.t. x1>=0.5( x1+x2+x3) +e
x3 <=0.1( x1+x2+x3) –e
y1>=0.2( x1+x2+x3) +e
y3 <=0.7( x1+x2+x3) –e
z1>=0.1( x1+x2+x3) +e
z3 <=0.5( x1+x2+x3) –e
x1+y1+z1 <=1500
x2+y2+z2 <=2000
x3+y3+z3 <=1000
x1,x2,x3,y1,y2,y3,z1,z2,z3>=0
Optimal Result:
x1=384.615
x2 =307.692
x3=76.923
y1=153.846
y2=1692.308
y3=538.462
z1=76.923
z2=0 . .
z3=384.615
w= 2676.923
补充:楼下说得不对,题目的要求是多于和少于,因此应该有e。
Max w=5.5(x1+x2+x3)+5(y1+y2+y3)+4.8(z1+z2+z3)-6(x1+y1+z1)-4.5(x2+y2+z2)-3(x3+y3+z3)
s.t. x1>=0.5( x1+x2+x3) +e
x3 <=0.1( x1+x2+x3) –e
y1>=0.2( x1+x2+x3) +e
y3 <=0.7( x1+x2+x3) –e
z1>=0.1( x1+x2+x3) +e
z3 <=0.5( x1+x2+x3) –e
x1+y1+z1 <=1500
x2+y2+z2 <=2000
x3+y3+z3 <=1000
x1,x2,x3,y1,y2,y3,z1,z2,z3>=0
Optimal Result:
x1=384.615
x2 =307.692
x3=76.923
y1=153.846
y2=1692.308
y3=538.462
z1=76.923
z2=0 . .
z3=384.615
w= 2676.923
补充:楼下说得不对,题目的要求是多于和少于,因此应该有e。
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