SOLUTION: Here is one of the hardest problems I have even seen! Can you please see if you can help with this one? I feel like I can write some functions, but I am really confused. Produ

Question 106606: Here is one of the hardest problems I have even seen! Can you please see if you can help with this one? I feel like I can write some functions, but I am really confused.
Product x: Each item sold produces $100 profit
Labor cost: is 3 hours for assembly, 1 hour in detail work, and 2 hours for QA & packing product.
Product y: Each item sold produces $150 profit
Labor cost: is 5 hours for assembly, 3 hour in detail work, and 2 hours for QA & packing product.
Company has these hours available for production for each accounting period:
3900 hours for assembly, 2100 hours for detail work, & 2200 hours for QA & packing.
Questions are:
1) How many products of both x & y should be produced to make a max & min profit?
2) Now what if product y profit goes up to $170 each. How many products are needed to make a max & min profit?
I know this is not an easy question, but do you think you could help with this question please? Thank you kindly.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Product x: Each item sold produces $100 profit
Labor cost: is 3 hours for assembly, 1 hour in detail work, and 2 hours for QA & packing product.
Product y: Each item sold produces $150 profit
Labor cost: is 5 hours for assembly, 3 hour in detail work, and 2 hours for QA & packing product.
Company has these hours available for production for each accounting period:
3900 hours for assembly, 2100 hours for detail work, & 2200 hours for QA & packing.
----------------
Profit function: Profit = 100x + 150y
Assembly : 3x+5y<=3900
Detail : x +3y<=2100
QA&Pack : 2x+2y<=2200
--------------------------
Questions are:
1) How many products of both x & y should be produced to make a max & min profit?
----------
You have to develop a solution set for the Assembly/Detail/QA&Pack inequalities
Then you have to determine the coordinates of the vertices of this solution
set.
Then you have to substitute the x/y values of each of these vertices into
the Profit function to see where you get a maximum profit and where you get
a minimum profit.
-------------------------------
2) Now what if product y profit goes up to $170 each. How many products are needed to make a max & min profit?
The only thing that will change is the Profit function; it will become
Profit = 100x + 170y
To find the man or min you follow the same procedure you did with the
former Profit equation.
==============================
This whole procedure is called linear programming.
================================
Cheers,
Stan H.

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