Question 818584: A chainsaw requires 3 hours of assembly and a wood chipper 5 hours. A maximum of 30 hours of assembly time is available. The profit is $130 on a chainsaw and $180 on a wood chipper. How many of each should be assembled to maximize profit?
I am not even sure where to start!
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! saw:
number produced: s
time to assemble: 3 hrs/unit
profit: 130 $/unit
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chipper:
number produced: c
time to assemble: 5 hrs/unit
profit: 180 $/unit
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3s + 5c = 30
3s = 30 - 5c
s = 10 - (5/3)c
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130s + 180c = p
130(10 - (5/3)c) + 180c = p
1300 - (650/3)c + 180c = p
1300 - (650/3)c + (540/3)c = p
1300 - (110/3)c = p
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profit turns out to be a linear function of the number of chippers produced !
p(c) = (-110/3)c + 1300
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use the slope-intercept form of the linear equation:
slope = (-110/3) = -36.666666
y-intercept = 1300
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plug this (the slope and y-intercept):
-36.666666 1300
into the "Slope–Intercept form: m b" input box here:
https://sooeet.com/math/linear-equation-solver.php
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x-intercept = 35.45
y-intercept = 1300
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answer:
profit is maximized at $1300 when zero chippers are produced, and profits go to zero when 35 chippers are produced.
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if zero chippers are produced, 10 saws are produced:
3s + 5c = 30
3s + 5(0) = 30
3s = 30
s = 10
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check:
profit when 10 saws are produced: (10 saw units) * (130 $/unit) = $1300 (correct)
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