Question

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?

Answer 1

ready ?

Answer 2

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

Answer 3

Ok

Answer 4

For one I don't know where to start

Answer 5

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

Answer 6

this is where x is the price of the item

Answer 7

now Profit = Sales - Cost

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

Answer 8

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

Answer 9

you can simplify this to


p(x) = 1000* {3-(x-6)}*(x-1)
= 1000* (9-x)*(x-1)

Answer 10

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.

Answer 11

sorry, (x-1) is actually (x-4)
I forgot to multiply 4 for the cost.

anyway, was that of any help ?

Answer 12

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

Answer 13

3000 is the y- intercept

Answer 14

-1000 is the slope

Answer 15

so when x=6, to make y = 3000

y has to be 3000-1000(x-6)

Answer 16

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

Answer 17

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

Answer 18

Yes

Answer 19

when x = 7, y = 2000

Answer 20

Yes

Answer 21

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

Answer 22

y - 3000 = -1000(x-6)

so y = 3000-1000(x-6)

Answer 23

Ok, I see it, lol

Answer 24

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

Answer 25

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)

Answer 26

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

Answer 27

woops, 13, not 4

Answer 28

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.

Answer 29

Ok, thanks for the help!!

Answer 30

good luck b