1.you can now sell 50 cups of lemonade per week at 40¢ per cup, but demand is dr
ID: 2902414 • Letter: 1
Question
1.you can now sell 50 cups of lemonade per week at 40¢ per cup, but demand is dropping at a rate of 5 cups per week each week. Assuming that raising the price does not affect demand, how fast do you have to raise your price if you want to keep your weekly revenue constant?
2. You can now sell 60 cars per month at $30,000 per car, and demand is increasing at a rate of 3 cars per month each month. What is the fastest you could drop your price before your monthly revenue starts to drop?
3.The demand equation for rubies at Royal Ruby Retailers is
q+1/2p=70
where q is the number of rubies RRR can sell per week at p dollars per ruby. RRR finds that the demand for its rubies is currently 15 rubies per week and is dropping at a rate of one ruby per week. How fast is the price changing? (Round your answer to the nearest cent.)
Explanation / Answer
Revenue = 50 cups * 50c = $25
It's not linear.
Required price in week n in cents
(where n = 0 in the initial 50 cups/week week)
=
2500 / (50 - 3*n)
Required increase = 2500 / (50 - 3(n-1)) - (2500 / (50 - 3n))
# sold ... price . . . . increase from previous week
50 ... 0.50 ...
47 ... 0.53 ... 0.03
44 ... 0.57 ... 0.04
41 ... 0.61 ... 0.04
38 ... 0.66 ... 0.05
35 ... 0.71 ... 0.06
32 ... 0.78 ... 0.07
29 ... 0.86 ... 0.08
26 ... 0.96 ... 0.10
23 ... 1.09 ... 0.13
20 ... 1.25 ... 0.16
17 ... 1.47 ... 0.22
14 ... 1.79 ... 0.32
11 ... 2.27 ... 0.49
8 ... 3.13 ... 0.85
5 ... 5.00 ... 1.88
2 ... 12.50 ... 7.50
Those $5.00 and $12.50 lemonades might prove a tough sell.
3) I am assuming the expression is q + (1/2)p = 60 Rewrite q + (1/2)p = 60 as:- (1/2)p = 60 - q p = 2(60-q) = 120 - 2q Differentiate p = (120-2q) with respect to time (t) dp/dt = (-2) x dq/dt Substitute dq/dt = -1 ruby per week ( minus sign because it is decreasing) dp/dt = (-2) x (-1) = 2 dollars per ruby per week = 200 cents per ruby per week Assuming you meant the expression as q + 1/(2p) = 60 Rewrite q + 1/2p = 60 as:- 1/2p = 60 - q p = 1/2(60-q) p = 1/(120-2q) Differentiate p = 1/(120-2q) with respect to time (t) dp/dt = [1/(120-2q)^2] x (-2) x dq/dt Substitute q = 25 and dq/dt = -1 ruby per week ( minus sign because it is decreasing) 1/4900 x (-2) x (-1) = 1/2450 dollars per week = 100/2450 cents per ruby per week = 1/24.5 cents per ruby per week
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