Quantitative Method for Business: max profit for glasswork

[darkdesign007]

I need help solving and setting up this problem. I have gotten the first 3 problems, but this #4 is really difficult for me. I need help to solve and set this problem up. Please any help would be appreciated.

4. A glass blower makes glass decanters and glass trays on a weekly basis. Work in progress cannot be carried over from week to week. Each item requires 1 pound of glass, and the glass blower has 15 pounds of glass available each week. A glass decanter requires 4 hours of labor, a glass tray requires 1 hour of labor, and the glass blower works 25 hours a week. The profit from a decanter is $50, and the profit from a tray is $10.

a. Find the optimal production schedule for this situation.

b. If the glass blower received an order for 5 trays that she could not refuse, what would be the optimal production schedule?

c. What would be the penalty for this additional constraint?

 

[stapel]

I have gotten the first 3 problems, but this #4 is really difficult for me. I need help to solve and set this problem up.

Since you completed the other exercises with no trouble, you'll have been able to make good progress on this one. So that we don't waste your time duplicating or explaining what you've already done, kindly please reply showing your work and reasoning so far, starting with which variables you chose and how you defined them.

Please be complete. Thank you!

Eliz.

 

[masters]

See if this makes any sense:

Assuming you will make at least 1 decanter and 1 tray, the first constraints would be d > 0 and t>0.

Next, since it takes 4 hrs of labor to make a decanter and 1 hr for a tray the constraint would be 4d + t <= 25

Then, since each require 1 lb of glass, a third constraint would be d + t <= 15

The max profit comes from 50d + 10t.


If you want to add the other constraint about a minimum of 5 trays, simply use t >= 5 instead of t > 0.

 

[masters]

Graph the inequalities and find the vertices of the feasible region.

I get (d,t) = (6.25, 0), (0, 0), (9/3, 35/3), and (0, 15)

Looks like production of 6 decanters and 0 trays yields a profit of $300.
Three decanters and 11 trays yields $260.
Zero decanters an 15 trays yields $150.

If you replace the t>=0 with t>=5, then a new vertex of (5, 5) yields a profit of 5 decanters and 5 trays for $300.