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anonymous

  • 5 years ago

A manufacturer has been selling flashlights at $6 a piece, and at this price consumers have been buying 3000 flashlights per month. The manufacturer wishes to raise the price and estimates that for each $1 increase in price, 1000 fewer flashlights will be sold each month. The manufacturer can produce the flashlight a cost of $4 per flashlight. At what price should the manufacturer sell the flashlights to generate the greatest profit?

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  1. Yuki
    • 5 years ago
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    ready ?

  2. Yuki
    • 5 years ago
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    i need to go soon so I will leave a couple of suggestions, ok ?

  3. anonymous
    • 5 years ago
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    Ok

  4. anonymous
    • 5 years ago
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    For one I don't know where to start

  5. Yuki
    • 5 years ago
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    the rate that the people buy this item is 3000 - 1000(x-6) since when the price is $6 , 3000 items are sold and for each $ increase from $6 the number of items sold decrease by 1000

  6. Yuki
    • 5 years ago
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    this is where x is the price of the item

  7. Yuki
    • 5 years ago
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    now Profit = Sales - Cost so we need to make a function of profit and maximize it. if we let Profit = p(x)

  8. Yuki
    • 5 years ago
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    p(x) = (number of items sold) * (price of each item) - (total cost for manufacturing the items) so p(x) = {3000-1000(x-6)} *x - (3000-1000(x-6)) i

  9. Yuki
    • 5 years ago
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    you can simplify this to p(x) = 1000* {3-(x-6)}*(x-1) = 1000* (9-x)*(x-1)

  10. Yuki
    • 5 years ago
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    to maximize this, you find p'(x) and find the local maximum. just in case check the boundary to see if the price was not the maximum.

  11. Yuki
    • 5 years ago
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    sorry, (x-1) is actually (x-4) I forgot to multiply 4 for the cost. anyway, was that of any help ?

  12. anonymous
    • 5 years ago
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    It does indeed help but I don't understand how you figure 3000-1000(x-6) is the rate of flashlights purchased

  13. Yuki
    • 5 years ago
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    3000 is the y- intercept

  14. Yuki
    • 5 years ago
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    -1000 is the slope

  15. Yuki
    • 5 years ago
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    so when x=6, to make y = 3000 y has to be 3000-1000(x-6)

  16. anonymous
    • 5 years ago
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    I don't know what it is about it, I just don't understand how you formulated that function.

  17. Yuki
    • 5 years ago
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    ok, when x =6, y = number of items sold = 3000 right ?

  18. anonymous
    • 5 years ago
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    Yes

  19. Yuki
    • 5 years ago
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    when x = 7, y = 2000

  20. anonymous
    • 5 years ago
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    Yes

  21. Yuki
    • 5 years ago
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    so, using \[y-y_1 = m(x-x_1)\]

  22. Yuki
    • 5 years ago
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    y - 3000 = -1000(x-6) so y = 3000-1000(x-6)

  23. anonymous
    • 5 years ago
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    Ok, I see it, lol

  24. anonymous
    • 5 years ago
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    So do I take 1000* (9-x)*(x-4) and expand it then differentiate ?

  25. Yuki
    • 5 years ago
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    if you don't mind the product rule of differentiation then you can just leave it like that. it does seem like it would be easier though. so yes, you let p'(x) = 0 and find the critical points. also check the endpoints because it is possible that when the price is cheapest or highest you get the maximum profit. (although in real life it doesn't happen)

  26. anonymous
    • 5 years ago
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    -2x + 4 = f'(x) right?

  27. anonymous
    • 5 years ago
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    woops, 13, not 4

  28. Yuki
    • 5 years ago
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    sorry I gotta go :/ but the differentiation part should be easy enough. I'm pretty sure you can get it :) I am guessing that the price has to be slightly higher than 6.

  29. anonymous
    • 5 years ago
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    Ok, thanks for the help!!

  30. Yuki
    • 5 years ago
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    good luck b

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