An electronics retailing chainhas established the monthlyprice (p)- demand (n )

Question

An electronics retailing chainhas established the monthlyprice (p)- demand (n ) relationship for a Nintendo game as: . They are trying to set a price levelthat will provide a maximum revenue (R). They know that when demandis elastic, a drop in price will result in higher overall revenuesand that when demand is inelastic, anincrease in price will result in higher overall revenues. Tocomplete the questions in this task, you will have to use theelasticity definition: E = - Converted into differential notation. Discussion Questions 1. Determine the elasticity ofdemand at $20 and $80, classifying these price points as havingelastic or inelastic demand. What does this say about where theoptimum price is in terms of generating the maximum revenue?Explain. Also calculate the revenue at the $20 and $80 pricepoints. ( 5 marks) 2. Determine the pricepoint where where you will not want to increase or decrease theprice to generate higher revenue. ( Hint: Think about the specificvalue of E where you won

Explanation / Answer

Kind of a strange thing to want(maximum revenue). I'd have thought maximum profit would be the way to go. You can plug in the values and see that when p is around 51, you get the maximum revenue and that if p=50 or p=52 you get lower values. p=50; R=25664.43448 p=51; R=25667.76801 p=52; R=25651.10154 Your elasticity value crosses the 1.0 boundary right about there too, though it actually crosses somewhere around $50.5 or something. That 1.0 value must be your special E value(or maybe I plugged in the numbers wrong). Using your differential version of E you see dE cross zero at that same spot. I suspect that if you read the book and use the information provided there will be a more exact way to determine the answer(less experimental).

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