A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit must be p=1000-x. The manufacturer also determines that the total cost of producing x units is given by C(x)=3000+20x. Find the total revenue R(x) and find the total profit P(x). How many units must the company produce and sell in order to maximize profit? What is the maximum profit?
Ok, so this is telliong us that is some one bought 25 stereos they woulb be charged 975 for the 25th one, it also tells us that the cost of producing An INFINITE amount of units is 3000 + 20 dollars more the stereo's you whant! so for example, if we whanted to make ...you still there?
yes thank you
ok, so, sorry it took me so long blarg. XD, so if we whanted tomanufacture 5 speakers it would only cost us 20 dollars more to make a new one.
now how do you think we can make the least amount of profit? :)
oh, 20 dollars plus the starting charge which is 3000 to start manufacturing.
How many units must the company produce and sell in order to maximize profit= how many do we have to sell so we dont get ripped off?
Yeah, which means I'm trying to figure out how many units of stereos, I'm probably over thinking this a lot, but how would I manipulate the available formulas to show me how to find the number of stereos?
It's quite easy, what you'r really looking for however is the line in which you start gaining profit, so, since we have a 3000 dollar starting charge, and the stereos cost 1000- the number of stereos you've sold.....
okay I think i'm starting to understand a bit.
Find the total revenue R(x) and find the total profit The revenue is the first thing we did so how much we earn by every stereo we sell.....the total profit is the total amount of money we have earned SO far, they are different formulas....now what im stuck in is to find the line were we start earning profit...
that's where i'm stuck too, what i posted is all the information i was given
p=1000-x the price per unit formula could also be used in the revenue formula, for example p=1000-x(100)....the outcome of that would be the profit earned, which we could write differently as, well, the same as above...
ok then, so far we have answered question 1, now to find when we can maximize our profits....
p=1000-x C(x)=3000+20x ok, we will kinda guesstimate in this scenario, I will post the chart and the outcomes....if we had 50 stereos....
and im pretty shure there's a formula for this to, but I really doubt you were suppose to find that....
@HyperChemist May I try?
Yes please....kinda stuck however, I cant understand if for the first one you have to subtract ALL the stereos, therefore giving you the ratio of 900:1 sterio or if the stereo's prices are different each time, decreasing by 1 dollar each time you sell one... or if the total amount , say, if you sold 30 stereos, you would have 970 dollars profit...hm...
yes, you may try, but I have come to the conclusion that if you buy 30 stereos, you can have them for 970 dollars....
yes, that is correct, however in real life inaccurate...
Ok, so, if we buy 100 stereos, we will have a 900 profit, and the manufactoring cost is 5000
no...I was correct, It says PRICE PER UNIT, SORRY FOR THE CONFUSION...
So yes! it's just a little discount that they give, 1- EVERY stereo that you buy for all of em!
ok, lets try that again.....we would actually have 9000 and 5000, resulting in a 4000 dollar profit for 100 stereos....
Revenue R(x) = x * P (x) = x ( -x + 1,000) = - x² + 1,000 x Profit P (x) = R(x) - C(x) = (- x² + 1,000 x ) - ( 20x + 3,000 ) = -x² + 980 x - 3,000 P'(x) = 0: - 2x + 980 = 0 -> x = 980/2 = 490 stereos =>Maximize Profit: P ( 490 ) = -490² + 980 (490) - 3,000 = ...
wow I totally get that now, any how... if we had 60 stereos.... the cost is 4200, and the profit is 56000, did some multiplication wrong the first time, but we're way closer now..
if we sold 10....9900 3200.....ok....
if we sold 3....
Douh! explain please!
0-0 wow I didnt know you responded! omg and wow I saw it now thank you so much!
definitely going to save this...huh..thanks!
I'm glad someone learns something =)
@AlexanderP759 Any questions?
Nope, that was exactly what I needed. Sorry I was absent, I was running an errand. Thank you very much :)
Don't worry, even you ran somewhere, but the post runs no where :P
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