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Quadratic Eqution word problem Help

 
 
Dyr
 
Reply Mon 12 Dec, 2011 08:33 pm
I really would appreciate help solving this equation. Step-by-step answers would also be thoroughly appreciated, too, if it isn't too much trouble. Thanks :3

the question ~
A theater seats 2000 people and charges 10$ for a ticket. At this price, all the tickets can be sold. A survey indicates that if the ticket price is increased, the number of tickets sold will descrease by 100 for every dollar of increase. What ticket price would result in the greatest revenue?
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Type: Question • Score: 0 • Views: 2,076 • Replies: 2
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markr
 
  1  
Reply Tue 13 Dec, 2011 12:45 pm
@Dyr,
tickets_sold = 2000 - 100*(cost-10)
revenue = tickets_sold * cost

Combine these to produce revenue as a function of cost, then determine which cost maximizes revenue by taking the derivative of that function and setting it to zero.
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raprap
 
  1  
Reply Wed 14 Dec, 2011 08:21 pm
@Dyr,
R=P(2000-100(P-10))=2000P-100 P^2+1000P=3000P-100P^2
dR/dP=3000-200P=0
200P=3000
P=3000/200=15
is this a max or min?
d2P/dR2=-200
so its a max

P=$15
R=3000*15-100*15^2=$22500

Rap
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