大学数学动态规划问题。某公司你投资500万到三个项目A、B、C，每个项目因为投资额度的不同获得的收益也不同，如图所示要求

fa(k) :invest k wanyuan on A project. fab(k) :invest k wanyuan on A and B projects
fabc(k) :invest k wanyuan on A,B and C projects.
step one
fabc(500) = max{ fc(500) , fc(400)+fab(100) , fc(300)+fab(200) ,fc(200) +fab(300) ,fc(100)+fab(400) ,fab(500) }
= max{ 110 ,90+fab(100) ,80+fab(200) 40 +fab(300) ,20+fab(400) ,fab(500) }.
step two
fab(500) = max{ fb(500) ,fb(400)+fa(100),fb(300)+fa(200),fb(200)+fa(300),fb(100)+fa(400),fa(500)}
= max{ 100,92+fa(100),70+fa(200),35+fa(300),16+fa(400),fa(500)}
= max{ 100,92+15,70+36,35+78,16+90,100}
= 35+78 = 113;
fab(400) = max{fb(400),fb(300)+fa(100),fb(200)+fa(200),fb(100)+fa(300),fa(400)}
=max {92,70+15,35+36,16+78,90}
= 16+78 = 94;
fab(300) = max{fb(300),fb(200)+fa(100),fb(100)+fa(200),fa(300)}
=max{70,35+15,16+36,78}
=78;
fab(200) = max{fb(200),fb(100)+fa(100),fa(200)}
= max{35,16+15,36};
=36;
fab(100) = max{16,15}
=16;
now back to step one for counting fabc(500)
fabc(500) = max{ 110 ,90+fab(100) ,80+fab(200) 40 +fab(300) ,20+fab(400) ,fab(500) }.
= max{110,90+16,80+36,40+78,20+94,113}
= 40+78 = 118 (this is the best profit)
Best wish for you.