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anonymous
 5 years ago
3. A video store rents 800 videos per week at $4/video. They did some research and found that for every $0.50 increase in rental price, they will rent 50 less videos.
a) Develop an equation that models the revenue, R, and number of price increases x.
b) What is the maximum revenue they can make
c) How much should they charge per video rental to maximise the revenue?
d) How many videos will they rent for maximum revenue
e) What will be their revenue if the rent for each video is $10 each?
anonymous
 5 years ago
3. A video store rents 800 videos per week at $4/video. They did some research and found that for every $0.50 increase in rental price, they will rent 50 less videos. a) Develop an equation that models the revenue, R, and number of price increases x. b) What is the maximum revenue they can make c) How much should they charge per video rental to maximise the revenue? d) How many videos will they rent for maximum revenue e) What will be their revenue if the rent for each video is $10 each?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the equation is Revenue = Price * Units Price = (4  .5x) Units = (800  50x) So when x increases price drops by increments of .5 and Units sold drops by 50 Multiply them and you have your equation Revenue = (4  .5x)(800  50x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Price should be (4+.5x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find where revenue is most take the derivative R(x) = 25x^2 +200x + 3200 R'(x) = 50x + 200 0 = 50x + 200 4 = x, so max at 4
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