Question 267753: A woodworking shop makes tables and chairs. To make a chair, 8 min is needed on the lathe and 8 min is needed on the sander. To make a table, 8 min is needed on the lathe and 20 min on the sander. The lathe operator works 6h/day and the sander operator works 7h/day. How many chairs and tables can they make in one day working at this capacity?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A woodworking shop makes tables and chairs.
To make a chair, 8 min is needed on the lathe and 8 min is needed on the sander.
To make a table, 8 min is needed on the lathe and 20 min on the sander.
The lathe operator works 6h/day and the sander operator works 7h/day.
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How many chairs and tables can they make in one day working at this capacity?
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Lathe Equation:: 8c + 8t = 6*60 minutes
Sander Equation: 8c +20t = 7*60 minutes
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Subtract 1st from 2nd to get:
12t = 60
t = 5 (# of tables that can be produced)
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Substitute into 8c+8t = 6*60 and solve "c":
8c + 8*5 = 6*60
8c = 320
c = 40 (# of chairs that can be produced)
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Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Let  represent the number of chairs made. Let  represent the number of tables made.
Then  is the number of minutes spent by a lathe operator making chairs and  is the number of minutes spent by a lathe operator making tables. The number of minutes available for the lathe operator is 6 times 60 equals 360 minutes. Hence:
Then is the number of minutes spent by a sander operator making chairs and  is the number of minutes spent by a sander operator making tables. The number of minutes available for the sander operator is 7 times 60 equals 420 minutes. Hence:
(Of course, all of the above presumes that neither the lathe operator or the sander operator takes a coffee break, goes to the restroom, smokes a cigarette, or goes to lunch for the entire 6 or 7 hour day. Apparently in the fantasy world of word problems, anything is possible.)
The factory cannot make less than zero chairs or less than zero tables. Hence:
and
Finally, both and must be integers. The factory won't make .47856 of a table.
420 divided by 20 is 21, so the most tables that can be made in a day is 21.
360 divided by 8 is 45, so the most chairs that can be made in a day is 45.
If you graph the set of 4 inequalities you will get an overlap trapezoidal shaped area that satisfies all 4 inequalities simultaneously. This is the area of feasibility. I'll leave it as an exercise for the student that the four vertices of this trapezoidal area are (0,0), (0,21), (40, 5), and (45, 0).
Which tells us that the factory could make nothing at all -- (0,0),
All tables, (0,21)
One possible mix of chairs and tables, (40, 5)
All chairs, (0, 45)
And several other combinations...
You could reduce the number of possibilities by insisting that for every table made, the factory has to make 4 chairs.
John
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