Observing children's needs and interests should affect the way you plan activities.
1 answer:
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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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