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Assume that the revenue function R and the cost function C for a business are given as: $$\begin{align*} R&=1163x−9x^2\\ C&=45x+22 \end{align*}$$ where $x$ is the daily production and sales (i.e., each item produced on a given day is sold on that day). Assume that business is currently such that 12 units are produced and sold per day and that the rate of change of production and selling is currently 1 units per day. Find the rate of change in revenue per day.

  • Find the rate of change in cost per day.
  • Find the profit function.
  • Find the rate of change or profit per day.
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It is considered very rude to simply post the statement of a problem (in the imperative, no less). You should state (i) Why you are considering this question (study for a test, self-study, review, homework, what?); (ii) What you've tried or why you are stuck or confused. –  Arturo Magidin May 1 '11 at 23:04
Please don't post in the imperative (i.e. ask a question, don't tell us what to do). What have you tried? –  Brian May 1 '11 at 23:05

1 Answer 1

If you have a revenue function and you want to find the rate of change in revenue per day, since your unit of time is in days, we can just compute the derivative of $R$ in terms of $t$ remembering to differentiate implicitly.

$$\frac{dR}{dt}=-18x\left(\frac{dx}{dt}\right) + 1163$$

Since we know that $\frac{dx}{dt} = 1$ and that $x=12$, we can find $\frac{dR}{dt}$ by simply substituting into our equation.

A similar situation occurs with your cost function, but is more obvious.

$$\frac{dC}{dt} = 45$$ By the definition of profit, the profit function is given by: $$P = R - C= (1163x - 9x^2) - (45x + 22) = -9x^2 + 1118x -22$$

Again, in order to find the rate of change, we interpret this as being the derivative, we differentiate $P$ in terms of $t$

$$\frac{dP}{dt} = -18x \left(\frac{dx}{dt}\right) + 1118$$

Again, we can substitute our known values for $\frac{dx}{dt}$ and $x$ in order to find the rate of change of profit per day.

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I suspect you're six months too late to help OP with his homework. But that's OK - Mathematics is eternal. –  Gerry Myerson Nov 22 '11 at 11:25
For anyone who happens to stumble across a similar problem in a google search, not to mentioned I haven't done one of these cost problems in a long time, so... Yet, Mathematics is eternal, and the internet is immortalized so who knows, it may be of use to some student on Mars in the year 2318 (who is in Grade 8 learning Calculus). –  Samuel Reid Nov 23 '11 at 7:19
If $\frac{dx}{dt}=1$ I know it doesn't make any difference, but it seems strange how $\frac{dx}{dt}$ is selectively included in some of the expressions. E.g. $\frac{d}{dt}(1163x−9x^2)=1163\frac{dx}{dt}-18x\frac{dx}{dt}$; I don't know why you kept one of the $\frac{dx}{dt}$s and not the other. –  Jonas Meyer Dec 22 '11 at 7:59
@Jonas: en.wikipedia.org/wiki/Chain_rule –  Samuel Reid Dec 22 '11 at 21:36
@SamuelReid: Thank you but I am familiar with the chain rule. See my comment for an instance of its use. Why I do not understand is your inconsistency in including the expression $\frac{dx}{dt}$, when its placement is as important (or not) in some of the places where it is omitted. But this is really unimportant, so if my comment is unclear I apologize. –  Jonas Meyer Dec 23 '11 at 3:57

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