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?

- anonymous

- jamiebookeater

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- yuki

ready ?

- yuki

i need to go soon so I will leave a couple of suggestions, ok ?

- anonymous

Ok

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## More answers

- anonymous

For one I don't know where to start

- yuki

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

- yuki

this is where x is the price of the item

- yuki

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

- yuki

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

- yuki

you can simplify this to
p(x) = 1000* {3-(x-6)}*(x-1)
= 1000* (9-x)*(x-1)

- yuki

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.

- yuki

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

- anonymous

It does indeed help but I don't understand how you figure 3000-1000(x-6) is the rate of flashlights purchased

- yuki

3000 is the y- intercept

- yuki

-1000 is the slope

- yuki

so when x=6, to make y = 3000
y has to be 3000-1000(x-6)

- anonymous

I don't know what it is about it, I just don't understand how you formulated that function.

- yuki

ok, when x =6, y = number of items sold = 3000 right ?

- anonymous

Yes

- yuki

when x = 7, y = 2000

- anonymous

Yes

- yuki

so, using \[y-y_1 = m(x-x_1)\]

- yuki

y - 3000 = -1000(x-6)
so y = 3000-1000(x-6)

- anonymous

Ok, I see it, lol

- anonymous

So do I take 1000* (9-x)*(x-4) and expand it then differentiate ?

- yuki

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)

- anonymous

-2x + 4 = f'(x) right?

- anonymous

woops, 13, not 4

- yuki

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.

- anonymous

Ok, thanks for the help!!

- yuki

good luck b

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