animallover
New member
- Joined
- Oct 4, 2013
- Messages
- 1
The first part of the problem is done: the question is A company manufactures three-speed, five-speed, and ten-speed bicycles. Each bicycle passes through three departments: fabrication, painting & plating, and final assembly. The relevant manufacturing data are given in the table. How many bicycles of each type should the company manufacture per day in order to maximize its profit? What is the maximum profit?
2) Identify the variables:
X=number of 3 speed bikes
Y=number of 5 speed bikes
Z=number of 10 speed bikes
i=maximize profit
3) The objective function and constraints in the correct format:
maximize i = 80x + 70y + 100z subject to
3x + 4y + 5z <= 120
5x + 3y + 5z <= 130
4x + 3y + 5z <= 120
4) The two solution lines correctly filled out:
Optimal Solution: i = 2500; x = 10, y = 10, z = 10 (solution exactly from online solver)
Rounded solution: i = 2500; x = 10, y = 10, z = 10 (rounded off solution)
That part is done, here is the next part which is done. The Deal:
Your bicycle manufacturing company is informed that the profit on the 10 speed can be doubled from $100 to $200 a unit if they make start to make the frame out of lighter materials. However your total costs are going to double if you produce the lighter frame.
This is the deal:
Your customer is offering to double your profit on the 10 speed but in return is requiring that you supply them with at least 6 units of the 3 and 5 speed units.
(add this constraint when considering the deal).
x + y >= 6
We had already figured out that it was not worth taking the deal by this solution: maximize i = 80x + 70y + 200z subject to
3x + 4y + 5z <= 120
5x + 3y + 5z <= 130
4x + 3y + 5z <= 120
x + y >= 6
Optimal Solution: i = 4410; x = 3, y = 3, z = 19.8 if we round it down to 19, i=4250 and i=5000 would have been worth taking the deal, so now I have to figure this out: If no then briefly explain how you calculated the break even profit on the 10 speed to make the deal work.
I'm not even sure where to start and apparently I'm getting no help, I tried politely asking but got rudely shunned by my professor. Didn't realize asking for help was a crime.
2) Identify the variables:
X=number of 3 speed bikes
Y=number of 5 speed bikes
Z=number of 10 speed bikes
i=maximize profit
3) The objective function and constraints in the correct format:
maximize i = 80x + 70y + 100z subject to
3x + 4y + 5z <= 120
5x + 3y + 5z <= 130
4x + 3y + 5z <= 120
4) The two solution lines correctly filled out:
Optimal Solution: i = 2500; x = 10, y = 10, z = 10 (solution exactly from online solver)
Rounded solution: i = 2500; x = 10, y = 10, z = 10 (rounded off solution)
That part is done, here is the next part which is done. The Deal:
Your bicycle manufacturing company is informed that the profit on the 10 speed can be doubled from $100 to $200 a unit if they make start to make the frame out of lighter materials. However your total costs are going to double if you produce the lighter frame.
This is the deal:
Your customer is offering to double your profit on the 10 speed but in return is requiring that you supply them with at least 6 units of the 3 and 5 speed units.
(add this constraint when considering the deal).
x + y >= 6
We had already figured out that it was not worth taking the deal by this solution: maximize i = 80x + 70y + 200z subject to
3x + 4y + 5z <= 120
5x + 3y + 5z <= 130
4x + 3y + 5z <= 120
x + y >= 6
Optimal Solution: i = 4410; x = 3, y = 3, z = 19.8 if we round it down to 19, i=4250 and i=5000 would have been worth taking the deal, so now I have to figure this out: If no then briefly explain how you calculated the break even profit on the 10 speed to make the deal work.
I'm not even sure where to start and apparently I'm getting no help, I tried politely asking but got rudely shunned by my professor. Didn't realize asking for help was a crime.