Music@life

Additional DetailsFormulate and then solve a linear programming model of this problem, to determine how many
containers of each product to produce tomorrow to maximize profits.

The company makes four
juice products using orange, grapefruit, and pineapple juice.
Product Retail Price Per Quart
Orange juice $1.00
Grapefruit juice .90
Pineapple juice .80
All-in-One 1.10
The All-in-One juice has equal parts of orange, grapefruit, and pineapple juice. Each product
is produced in a one- quart size. On hand are 400 gallons of orange juice,300 gallons of grapefruit
juice, and 200 gallons of pineapple juice.

The cost per gallon is $2.00 for orange juice,
$1.60 for grapefruit juice,
and 1.40 for pineapple juice.
In addition, the manager wants grapefruit juice to be used for no more than 30 percent of
the number of containers produced( grapefruit <= 0.3(orange +grapefruit + pineapple + all-in-one). She wants the ratio of the number of containers of orange
juice to the number of containers of pineapple juice to be at least 7 to 5.( orange:pineapple >= 7:5)


1) find how many containers of each product to produce tomorrow to mazimize profits.

2) Find any Surplus and/or Slack.

I've been looking at this for days.. and I can't seem to grasp the idea and work it out.

*Notes:

grapefruit juice to be used for no more than 30 percent of

the number of containers produced
 ==>
     grapefruit <= 0.3(orange +grapefruit + pineapple + all-in-one).
因為他說no more than 所以是"<="
"the number of containers produced"是指所有飲料(美國的果汁都用像是台灣大桶牛奶塑膠罐裝)製造數量 , 30percent(0.30),所以是0.3(orange +grapefruit + pineapple + all-in-one).
我上次答案忘了加上all-in-one
 
 the ratio of the number of containers of orange
juice to the number of containers of pineapple juice to be at least 7 to 5.
==>
( orange:pineapple >= 7:5)
因為他說 at least 所以是 ">="
 
所有變數都得移項到左邊，右邊(RHS)只有數值
就不演繹
這是我最後確認的答案 :-(

ANS:

Let orange juice as X1, grapefruit juice as X2, pinapple juice as X3, all-in-one as X4

Run following fomulas in LINDO:

Max 0.5X1+0.5X2+0.45X3+0.68X4
ST
1X1+0.3333X4<=1600
1X2+0.3333X4<= 1200
1X3+0.3333X4<=800
0.7X2-0.3X1-0.3X3-0.3X4<=0
5X1-7X3 >=0
END  Global optimal solution found.
  Objective value:                              2232.163
  Infeasibilities:                              0.000000
  Total solver iterations:                             5

                       Variable           Value        Reduced Cost
                             X1        800.0000            0.000000
                             X2        400.0000            0.000000
                             X3        0.000000           0.5902040
                             X4        2400.240            0.000000

                            Row    Slack or Surplus      Dual Price
                              1        2232.163            1.000000
                              2        0.000000           0.5000000
                              3        0.000000           0.5000000
                              4        0.000000            1.040204
                              5        680.0720            0.000000
                              6        4000.000            0.000000