Question 62157: A dreaded word problem. I can't figure this one out.
The Municipal transit serves 150,000 commuters daily when the fair is $1.80.
Market research has determined that every penny decrease in the fare will result in 1,000 new riders. What fare will maxmize revenue?
Any help would be appreciated. Thanks!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The Municipal transit serves 150,000 commuters daily when the fair is $1.80.
Market research has determined that every penny decrease in the fare will result in 1,000 new riders. What fare will maxmize revenue?
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Look at it a penny at a time and try to see the pattern:
If Fare is 180 cents the # of commuters is 150,000
If Fare is 180-1 cents the # of commuters is 150,000+1000
If Fare is 180-2 cents the # of commuters is 150,000+2(1000)
...
If Fare is 180-x cents the # of commuters is 150,000+x(1000)
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EQUATION:
Revenue = (Fare)(# of commuters)
R(x)=(180-x)(150,000+1000x)
R(x)= -1000x^2-150000x+180000x+180*150000
R(x)=-1000x^2+30000x+180*150000
Maximum occurs at x=-b/2a = -30000/-2000=15
Maximum Revenue comes when Fare = 180-15=165 cent or $1.65
Cheers,
Stan H.
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