（2）最小权和为18
八、解：设产品甲、乙的产量分别为x1,x2，
??minz?p1d1??p2(d2?d2)?p3d3??p4(?6x1?9x2)?20x1?30x2?1000?6x?9x?d??d??3001211????s..t?x1?1.5x2?d2?d2?0?30x?50x?d??d??1500233?1????xi?0,i?1,2.dj,dj?0,j?1,2,3九、解：设产品甲、乙的产量分别为x1,x2，
?????minp1d1??p2(d2?d2)?p3(d3?d3)?p4d4
?3x1?12?4x?16?2?20x1?40x2?d1??d1??80???s.t.?x1?x2?d2?d2?0?2x?2x?d??d??12233?1???d4?15?5x1?3x2?d4????x1,x2,di,di?0,i?1,2,3,4 十、选方案a3
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十一、解：假设 x ij ,i ? 1 j ? 1 ,2,3,4 表示第i个产粮地运往第j个需求地,2,3;的运量（万吨）用Z表示总运输费用,则得:
minZ/10000?3x11?2x12?6x13?3x14
?5x21?3x22?8x23?2x24?4x31?x32?2x33?9x34
?x11?x12?x13?x14?10 ?x?x?x?x?8?21222324
?x31?x32?x33?x34?5
? ?x11?x21?x31?5st:? ?x12?x22?x32?7 ?x13?x23?x33?8?
?x14?x24?x34?3 ?x?0,i?1,2,3j?1,2,3,4?ij
LINGO程序： model: sets:
chandi/1..3/:chanliang; xiaodi/1..4/:xuqiu;
yunfei(chandi,xiaodi):c,x; endsets data:
chanliang=10,8,5;
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xuqiu=5,7,8,3; c=3,2,6,3 5,3,8,2 4,1,2,9; enddata
min=@sum(yunfei:c*x);
@for(chandi(i):@sum(xiaodi(j):x(i,j))= chanliang(i)); @for(xiaodi(j):@sum(chandi(i):x(i,j))= xuqiu(j)); end
dx(t)?kx(t) 十二、解：设t时刻司机血液中酒精含量为x(t)（mg/ml），建立微分方程模型：dtdx?kdt,x?cekt 求解模型：x据已知条件可得：0.56?ce3k,0.4?ce5k 由此解得k??0.1682,c?0.9275
故x(t)与时间t的函数关系式为x(t)?0.9275e?0.1682t，因为x(0)?0.9275?0.80所以可判定该司机为醉酒驾车。
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