Looking for some math help

[Acevalefor]

You failed to math.

.9x + 40 = .7x + 300
.2x = 260
x = 1300

Thanks. Math fail indeed.

Thanks for the website quannum.

[Aylee]

Alright new question.

The maringal cost of producting the xth roll of film is 5 + 2x + 1/x. The toal cost to produce one roll is $1000. Find the Cost function C(x).

From what I could understand from youtube videos is you just plug $1000 in for X but that seems way to fucking simple to be right. Am I missing something or is it really that easy?

[Byrthnoth]

You are missing something, if I understand the question right.

They probably want the cost of producing an arbitrary number of films.

You solve the given equation for x=1 to get your offset. 5+2+1=8, but they are telling you it cost 1000 so there must be $992 of fixed costs inherent in making film.

Now you need to add up all the costs of each roll.

Edit I started trying to drive a more general form, but this is probably what they are looking for:
992 + sigma(5 + 2x + 1/x) from 1 to x

Are you doing Taylor series and shit in class?

[Blubbartron]

The marginal cost function is the derivative of the cost function. If you're given the marginal cost function, find the antiderivative, using the offset that Byrthnoth was talking about to determine the value of C (the unknown constant of any antiderivative).

http://www.dummies.com/how-to/conten...evenue-an.html

[Salodin]

It wasn't until this year that I realized how awesome calculus is.

[Aylee]

You are missing something, if I understand the question right.

They probably want the cost of producing an arbitrary number of films.

You solve the given equation for x=1 to get your offset. 5+2+1=8, but they are telling you it cost 1000 so there must be $992 of fixed costs inherent in making film.

Now you need to add up all the costs of each roll.

Edit I started trying to drive a more general form, but this is probably what they are looking for:
992 + sigma(5 + 2x + 1/x) from 1 to x

Are you doing Taylor series and shit in class?

no clue what the taylor series is so i guess not lol. went over the question in class and we had to integrate it, then plug in 1000 for x. so it was that simple i was just missing the integration part.

[Blubbartron]

no clue what the taylor series is so i guess not lol. went over the question in class and we had to integrate it, then plug in 1000 for x. so it was that simple i was just missing the integration part.

Out of curiosity, did you not notice my post immediately after Byrthnoth's, or are they no longer teaching "antiderivative" as the proper term these days? Usually people only refer to it as an integral when it's a definite integral (i.e. has proper lower/upper bounds so that the integration resolves to a number).

[BerenTebogo]

Out of curiosity, did you not notice my post immediately after Byrthnoth's, or are they no longer teaching "antiderivative" as the proper term these days? Usually people only refer to it as an integral when it's a definite integral (i.e. has proper lower/upper bounds so that the integration resolves to a number).

Antiderivative and indefinite integral are interchangeable and everyone who uses them understands they both mean the same thing. Calling attention to the difference really isn't necessary.

[Blubbartron]

Antiderivative and indefinite integral are interchangeable and everyone who uses them understands they both mean the same thing. Calling attention to the difference really isn't necessary.

I agree, but he didn't use the keyword you did (bolded). I was just curious if it was a gap in vocabulary or what, because I'm used to only seeing/using "integrate" and "integral" when dealing with the definite version. My professors were a bit old fashioned and rigid, so I wouldn't be shocked if things have changed in the decade or so since I took calculus.

[Aylee]

Sorry I suck at what shits called I just know what they look like lol. We were integrating some and doing definite integrals on others (which really fucked me up cuz i'd always forget the +C on the integration). So we did the integrate like ∫ 2x + 12 and then we were doing the same thing on other problems but with the upper and lower bounds.