Question 658996: A furniture shop refurnishes tables. Employees use 1 of two methods to refinish each table. Method I takes 1 hour to finish and the material costs $7. method II takes 1.5 hours, and the material costs $6. Next week, they plan to spend 88 hours in labor and $526 in material for refinishing tables. How many tables should they plan to refinish with each method?
Found 2 solutions by jim_thompson5910, Theo:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let x = # of tables made with method 1, y = # of tables made with method 2
Hours:
1x+1.5y = 88
x = 88 - 1.5y
Cost:
7x+6y = 526
7(88 - 1.5y)+6y = 526
7(88-15y)+6y=526
616-10.5y+6y=526
-4.5y+616=526
-4.5y=526-616
-4.5y=-90
y=(-90)/(-4.5)
y=20
x = 88 - 1.5y
x = 88 - 1.5(20)
x = 58
So you should make 58 tables using method 1 and 20 tables using method 2
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! method 1 takes 1 hour to finish and material costs $7.00
method 2 takes 1.5 hours to finish and the material costs $6.00
next week, they plan to spend 88 hours in labor and $526 in material.
it looks like the intersection points of interest are (0,58),(58,20),(75,0)
the total dollars and hours at these intersection points are:
(0,58) = 0 method 1 tables and 58 method 2 tables for a total material cost of 58 * 6 = $348 in material and 58 * 1.5 = 87 hours of labor.
(58,20) = 58 method 1 tables and 20 method 2 tables for a total material cost of 58 * 7 + 20 * 6 = $526 material cost and total labor of 58*1 + 20*1.5 = 88 hours of labor.
(75,0) = 75 method 1 tables and 0 method 2 tables for a total material cost of 75 * 7 = $525 and total labor of 75*1 = 75 hours.
the plan that appears to be the most efficient that makes maximum utilization of the material money available and the hours of labor available is the one that produces 58 method 1 tables and 20 method 2 tables.
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