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. CAN YOU

Question 106877: 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.
CAN YOU PLEASE HELP ME WORK THIS PROBLEM OUT IN FULL PLEASE?
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 ankor@dixie-net.com(22740) (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.
:
Assembly:
3x + 5y =< 3900
5y =< -3x + 3900; arrange in the "y=" form for graphing
y =< -.6x + 780; divided equation by 5
:
Detail
1x + 3y =< 2100
3y =< -x + 2100
y =< .33x + 700; divided equation by 3
:
QA & Packing
2x + 2y =< 2200
2y =< -2x + 2200
y =< -x + 1100; divided equation by 2
:
Graph this and you should have:

Note that the area of feasibility is at or below the the lowest graphing line
:
Questions are:
1) How many products of both x & y should be produced to make a max & min profit?
:
Find the coordinates of the corners of the area of feasibility, use these to
make an objective profit function for each corner.
The x/y values of the corners will be:
0, 700
300, 600
800, 300
1100, 0
:
0,700:
100x + 150y =
100(0) + 150(700) = $105,000 profit
:
300,600
100(300) + 150(600) =
30000 + 90000 = $120,000 profit
:
800,300
100(800) + 150(300) =
80000 + 45000 = $125,000 profit
:
1100, 0
100(1100) + 150(0) = $110,000 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 am pressed for time, here. Substitute 170 for 150 and do the same as
above. If you have any questions email me, I will respond tonight.
I know this is not an easy question, but do you think you could help with this question please? Thank you kindly.
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