EBike, an electronic bicycle manufacturer, has identified two customer segments, one is willing to pay a higher price for a cust
omized bicycle and the other one is more price sensitive and is willing to pay for standard bicycles. Demand curve for the customers willing to pay for customized bicycles is d1 = 11,000 - 25p1. Demand curve for the more price sensitive customers is d2 = 11,000 - 45p2. Assume production cost for each bicycle (standard and customized) is c = $160 per unit.
1. What price should eBike charge each segment if its goal is to maximize profits?
2. If eBike was to charge a single price over both segments, what should it be? How much increase in profits does differential pricing provide?
3. If total production capacity is limited to 5,000 units, what should eBike charge each segment?
1. What price should eBike charge each segment if its goal is to maximize profits?
2. If eBike was to charge a single price over both segments, what should it be? How much increase in profits does differential pricing provide?
3. If total production capacity is limited to 5,000 units, what should eBike charge each segment?
1 answer:
Solution:
1)
Profit function of customised bicycle customers, P1 = d1*(p1-c)
= (11000-25p1)*(p1-160)
= 11000p1-1760000-25p1^2+4000p1
= -25p1^2+15000p1-1760000
In order to the profit maximizing price, equate the first order derivative of profit function to 0
dP1/dp1 = d(-25p1^2+15000p1-1760000)/dp1 = 0
=> -50p1+15000 = 0
=> p1 = 300
Profit function of price sensitive customers, P2 = d2*(p2-c)
= (11000-45p2)*(p2-160)
= 11000p2-1760000-45p2^2+7200p2
= -45p2^2+18200p2-1760000
In order to the profit maximizing price, equate the first order derivative of profit function to 0
dP2/dp2 = d(-45p2^2+18200p2-1760000)/dp2 = 0
=> -90p2+18200 = 0
=> p2 = 202.22
Price to be charged for customised segment = $ 300
Price to be charged for price sensitive segment = $ 202.22
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2)
Considering single price, p
Total profit from both segments, P = (d1+d2)*(p-160)
= (11000-25p+11000-45p)*(p-160)
= (22000-70p)*(p-160)
= 22000p-3520000-70p^2+11200p
= -70p^2+33200p-3520000
In order to the profit maximizing price, equate the first order derivative of profit function to 0
dP/dp = d(-70p^2+33200p-3520000)/dp = 0
=> -140p+33200 = 0
=> p2 = 237.14
3)
This is solved by Solver as follows:
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Answer:
$2500
Explanation:
Given: Total output(quantity)= 1000 units.
average variable cost per unit= $3
Average fixed cost per unit= $1.5
Selling price per unit is $7
We know, Profit=
First, lets find out total cost
Total cost=
Remember, cost=
∴ Total cost= =
Now, finding total revenue.
Profit=
∴ Total profit=
Total profit= $2500