1. A factory manufactures two products, each requiring the use of three machines. The first machine can be used at most 60 hours; the second machine at most 30 hours; and the third machine at most 80 hours. The first product requires 2 hours on machine 1, 1 hour on machine 2, and 1 hour on machine 3; the second product requires 1 hours each on machine 1 and 2, and 3 hours on machine 3. If the profit is RM30 per unit for the first product and RM50 per unit for the second product, how many units of each product should be manufactured to maximize profit?
just noticed an arithmetic error..
in my evaluation of y, should be
60 - 2x
= 60-2(20)
= 20
so .....
max profit is 30(20) + 50(20) or 1600 , when 20 of the first product and 20 of the second product are made
Question Number 1. A company manufactures two products X1 and X2 on three machines A, B, and C. X1 require 1 hour on machine A and 1 hour on machine B and yields a revenue of Birr 3 Product X2 requires 2 hours on machine A and 1 hour on machine B and 1 hour on machine C and yields revenue of Birr 5. In the coming planning period the available time of three machines A, B, and C are 2000 hours, 1500 hours and 600 hours respectively.
let the number of the first product be x
let the number of the 2nd product by y
So we need to look at the regions
2x+y ≤60 and
x+y≤30 and
x+3y≤80
The profit function is 30x+50y
which has a slope of -3/5
move that profit line to the right, parallel to itself, until you reach the farthest vertex of that region.
From my rough sketch that appears to be the
intersection of
2x+y = 60 , ---> y = 60-2x
and
x + 3y = 80
substitution:
x + 3(60-2x) = 80
x-6x = -100
x = 20
then y = 60-20 = 40
max profit is 30(20) + 50(40) or 2600 , when 20 of the first product and 40 of the second product are made
Check: intersect the other two lines for the other vertex
x+y=30
x+3y=80
subtract them
2y = 50
y = 25 , then x = 5
profit = 30(5) + 50(25) = only 1400
To maximize profit, we need to determine the number of units to manufacture for each product. Let's assume we manufacture 'x' units of the first product and 'y' units of the second product.
The constraints given are:
- The first machine can be used at most 60 hours.
- The second machine can be used at most 30 hours.
- The third machine can be used at most 80 hours.
For the first product, the time required on each machine is:
- 2 hours on machine 1
- 1 hour on machine 2
- 1 hour on machine 3
So, the total time required for 'x' units of the first product on each machine is:
- Machine 1: 2x hours
- Machine 2: x hours
- Machine 3: x hours
Similarly, for the second product, the time required on each machine is:
- 1 hour on machine 1
- 1 hour on machine 2
- 3 hours on machine 3
So, the total time required for 'y' units of the second product on each machine is:
- Machine 1: y hours
- Machine 2: y hours
- Machine 3: 3y hours
Now, let's write the constraints using the given time limits:
- Machine 1: 2x + y ≤ 60
- Machine 2: x + y ≤ 30
- Machine 3: x + 3y ≤ 80
Next, we need to determine the objective function to maximize profit.
The profit for the first product is RM30 per unit, so the total profit for 'x' units of the first product is 30x.
The profit for the second product is RM50 per unit, so the total profit for 'y' units of the second product is 50y.
Therefore, the objective function is:
Profit = 30x + 50y
To summarize, we have the following equations and constraints:
Equations:
- Machine 1: 2x + y ≤ 60
- Machine 2: x + y ≤ 30
- Machine 3: x + 3y ≤ 80
Objective function:
Profit = 30x + 50y
Now, we can solve this problem using linear programming techniques, such as graphical method or simplex method, to find the values of 'x' and 'y' that maximize the profit.