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
