Answer:
max z =
Constraints
(Grade 9 oranges)
(Grade 6 oranges)
(Avg oranges in bags)
(Avg oranges in juice)
Explanation:
Let the variables used be x and y
x representing oranges of grade 9
y representing oranges of grade 6
Now, let be the oranges used in each bag in lbs, and be oranges used in juice of grade 9 each.
Similarly let will represent oranges of grade 6 in bags, and will represent oranges in juice of grade 6
Now total oranges sold in bags
= +
And their revenue in $ = 0.5 revenue - 0.2 expense = 0.3
Total profit from bag shall be
0.3 () = 0.3 + 0.3
Similarly total oranges in juice shall be
=
Profit shall be
$1.50 - $1.05 = $0.45
Total profit from juice shall be
$0.45 () =
Profit shall maximise as
z =
Further constraint 1 shall be
Total amount of grade 9 oranges used shall be max 100,000 lb
Constraint 2
Total amount of grade 6 oranges used shall be max 120,000 lb
Constraint 3
Average quality of oranges sold in bag shall be 7
Accordingly,
Simplifying:
Constraint 4
Average quality of oranges sold as juice shall be 8
Accordingly,
Simplifying:
Constraints shall be