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- This topic has 3 replies, 2 voices, and was last updated 4 years ago by John Moffat.
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- August 20, 2020 at 3:29 pm #581301
Hi John, I hope you are well.
I was wondering if you could help me with a question, I’m having trouble understanding how to work out this type of question. The question is –
What is the optimum selling price at the profit maximising level of sales?
They provide the following information in the question –
Market research has revealed that the maximum demand for Jewel Co’s earphones in the USA is 72,000 units per year, and that demand will reduce by 8000 units for every $5 that the selling price is increased. Jewel Co has calculated the profit maximising level of sales for its earphones for the coming year is 32,000 units.From my understanding we have to use the price demand equation to work this out. I was able to get as far as working out that b=5/8000 and that Q would be 32,000. I was unsure how to continue from here as I couldn’t seem to work out what ‘a’ would be. I understand the first 2 lines of kaplan’s solution below but then I’m confused as to how they worked out the remainder.
The solution in the kaplan exam kit is as follows:
P=0, demand (Q)=72,000 units
P=5, Q=72,000 – 8000 =64,000 units
Q=72,000-5P (P is the selling price in $ because demand will drop by 8000 units for every $5 increase in the selling price)
If the optimum quantity Q=32,000 units, P=5/8000 (72000-32000units) = $25.Please could you kindly help me understand how to work this question out? Please let me know if any further information is needed as the detail to the question was quite lengthy but I have included what I thought would be relevant.
Thank you in advance π
BhavniAugust 20, 2020 at 4:52 pm #581312Here we go π
P = a – bQ
The lower the selling price is then the higher the demand will be. Maximum demand is when the selling price (P) is zero, and this is 72,000 units.
Therefore 0 = a – 5/8,000 x 72.000 = a – 45
So a = 45The price/demand equation is therefore P = 45 – (5/8,000)Q
When Q = 32,000, P = 45 – 20 = $25
Have you watched my free lectures on this? The lectures are a complete free course for Paper PM and cover everything needed to be able to pass the exam well π
September 16, 2020 at 9:27 pm #585866Hi John, thank you so much for taking the time out to explain this to me, I really appreciate it. I did watch the lectures and use the notes alongside watching them. Both were extremely helpful and really did help me understand PM. I took the exam last week, fingers crossed it went well. Thank you again, Bhavni π
September 17, 2020 at 11:11 am #585904I am pleased that the exam went well for you π
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