SOLUTION: The problem is:
A DVD company wants to maximize its revenue. the company now sells dvd's for $40 each and they have 6000 customers. a study suggests that for every $2.00 increase
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-> SOLUTION: The problem is:
A DVD company wants to maximize its revenue. the company now sells dvd's for $40 each and they have 6000 customers. a study suggests that for every $2.00 increase
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Question 25817: The problem is:
A DVD company wants to maximize its revenue. the company now sells dvd's for $40 each and they have 6000 customers. a study suggests that for every $2.00 increase in price they will lose 200 customers. what should be the selling price to maximize the company's revenue? what will that revenue be?
So far I've written an equation:
let R= revenue
let x= # of $2 increases
R=(6000-200x)(40+x)
When i try to solve, I get: x=-5 and R=245000 but I know x can not be negative. Answer by Paul(988) (Show Source):
You can put this solution on YOUR website! Let x be the $ 2 increase.
cost of DVD:
2(x+40)
2x+80
DVD sold = 6000-200x
(2x+80)(6000-200x)
-400x^2+12000x+480000-16000x
-400x^2-4000x+480000
400x^2+4000x-480000=0,x
x^2+40x-1200=0
CAN YOU SOLVE IT FROM ZTHERE if not e-mail me back.
you'll get x=30
Hence, the revenue is 30.
PAUL.
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