Respuesta :
Answer:
max z = [tex]0.3x_{1} + 0.45x_{2} + 0.3y_{1} + 0.45y_{2}[/tex]
Constraints
[tex]x_{1} + x_{2} \leq 100,000[/tex] (Grade 9 oranges)
[tex]y_{1} + y_{2} \leq 120,000[/tex] (Grade 6 oranges)
[tex]2x_{1} - y_{1} \geq 0[/tex] (Avg oranges in bags)
[tex]x_{2} - 2y_{2} \geq 0[/tex] (Avg oranges in juice)
[tex]x_{1}\geq 0, x_{2} \geq 0, y_{1}\geq 0, y_{2} \geq 0[/tex]
Explanation:
Let the variables used be x and y
x representing oranges of grade 9
y representing oranges of grade 6
Now, let [tex]x_{1}[/tex] be the oranges used in each bag in lbs, and [tex]x_{2}[/tex] be oranges used in juice of grade 9 each.
Similarly let [tex]y_{1}[/tex] will represent oranges of grade 6 in bags, and [tex]y_{2}[/tex] will represent oranges in juice of grade 6
Now total oranges sold in bags
= [tex]x_{1}[/tex] + [tex]y_{1}[/tex]
And their revenue in $ = 0.5 revenue - 0.2 expense = 0.3
Total profit from bag shall be
0.3 ([tex]x_{1} + y_{1}[/tex]) = 0.3 [tex]x_{1}[/tex] + 0.3[tex]y_{1}[/tex]
Similarly total oranges in juice shall be
= [tex]x_{2} + y_{2}[/tex]
Profit shall be
$1.50 - $1.05 = $0.45
Total profit from juice shall be
$0.45 ([tex]x_{1} + y_{1}[/tex]) = [tex]0.45 x_{2} + 0.45 y_{2}[/tex]
Profit shall maximise as
z = [tex]0.3x_{1} + 0.45x_{2} + 0.3y_{1} + 0.45y_{2}[/tex]
Further constraint 1 shall be
Total amount of grade 9 oranges used shall be max 100,000 lb
[tex]x_{1} + x_{2} \leq 100,000[/tex]
Constraint 2
Total amount of grade 6 oranges used shall be max 120,000 lb
[tex]y_{1} + y_{2} \leq 120,000[/tex]
Constraint 3
Average quality of oranges sold in bag shall be 7
[tex]\frac{9x_{1} + 6y_{1} }{x_{1} + y_{1} } \geq 7[/tex]
Accordingly,
[tex]9x_{1} + 6y_{1} \geq 7x_{1} + 7y_{1}[/tex]
Simplifying:
[tex]2x_{1} - y_{1} \geq 0[/tex]
Constraint 4
Average quality of oranges sold as juice shall be 8
[tex]\frac{9x_{2} + 6y_{2} }{x_{2} + y_{2} } \geq 8[/tex]
Accordingly,
[tex]9x_{2} + 6y_{2} \geq 8x_{2} + 8y_{2}[/tex]
Simplifying:
[tex]x_{2} - 2y_{2} \geq 0[/tex]
Constraints shall be
[tex]x_{1}\geq 0\\x_{2} \geq 0\\y_{1}\geq 0\\y_{2} \geq 0[/tex]