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PLEASE help. idk how to maximize revenue 🙁

the environmental club sells sweatshirts. they sell 1200 shirts a year at ![20 each. they are planning to increase the price. a survey indicates that for every 20 each. they are planning to increase the price. a survey indicates that for every2 increase in price, there will be a drop of 60 sales a year. what should the selling price be in order to maximize the revenue?

So we have the information from the survey that for every extra 2 dollars 60 fewer shirts are sold. Alternatively, you could say that every extra dollar means that 30 fewer shirts are sold. So for every dollar increase in x we need to subtract thirty from the number of shirts sold. You can write this as: Shirts sold = 1200 - 30(p-20) Where p is the new price of the shirts. You can see that if we didn't change the price at all, ie p = 20, then 20 - 20 = 0 and so the -30(p-20) part of this vanishes, so that we are left with selling 200 shirts as before. Now rearrange this a bit: 1200 - 30(p-20) = 1200 - 30p - 30 x (-20) = 1200 - 30p + 600 = 1800 - 30p It's also helpful for later, to reduce the size of the numbers we are working with, to notice that 1800 = 30 x 60 so we can write: Shirts sold = 30(60 - p) Now to find out how much revenue you make, multiply the number of shirts sold by their price, with is p, to get: Revenue = 30p(60-p) Or, cheating a little bit again to reduce the size of the numbers we are talking about (but DO NOT FORGET that you have done this!): Revenue/30 = p(60-p) Here I've divided both sides of the equation by 30. The argument is that whenever the revenue is maximised any fixed fraction of it will be maximised too. Which hopefully you can see must be true. Getting rid of the 30 just leaves smaller numbers to deal with and saves on a bit of working later. In the question you've asked we don't even need to find the exact revenue anyway, so getting ridding of the 30 in this way is perfectly sensible. But if you were asked to find the revenue then you would have to remember at the end to remultiply by 30. Anyway, what we are left with is to find the maximum value of the following equation: y = p(60-p) = 60p - p^2 Which is a quadratic, and so the maximum can by found by standard differentiating techniques. You should check for yourself that you can recover the answer p = 30 for yourself (which gives a maximum revenue of $27000). Hope this helps.
29 October 2012
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