I realize that this problem is probably very trivial to most of you all here. But, this is a problem from a Calculus 1 class, and I was wondering what answer you all would get. I end up getting a selling price of $9.17.
Anne sells candles from her website. She can get a candle from her suppliers at a cost of $\$3$ to her. The candles have been selling for $\$5$, and at this price, consumers have been buying a steady $4000$ per month. Beth would like to raise prices, but a market surveyor told her that for each increase of $\$1$ in the price of an candle, she would lose $300$ more sales each month. At what price should she sell the candles to maximize her profit?
Now, I know that Profit = Revenue - Cost. For revenue, I calculated, $p∗[4000−300(p−5)]$, where $p$ is the amount of money that Anne should sell her candle. So, $p*[4000−300(p−5)]$ should be revenue. But then, for cost, do I also have to set cost = $3∗[4000−300(p−5)]$, and then solve for $p$, the price she should sell her candles, to maximize her profit?
You can do this entirely with algebra.
Profit = units * price - units * cost
$P = (4000-300(p-5))p - (4000-300(p-5))3\\ P = ap^2 + bp + c$
that is a parabola and we have a formula for the vertex.
$p = \frac {b}{2a}$
You can get fancy and use calculus. You can differentiate the same equation find where $\frac {dP}{dp} = 0$
A handy one for economics is to know "marginal revenue equals marginal costs."
That is:
Profit = Revenue - Costs
$\frac {d}{dq} P = \frac {d}{dq} R - \frac {d}{dq} C = 0\\ \frac {d}{dq} R = \frac {d}{dq} C$
Note that I am differentiating with respect to units and not prices. And unit cost is not changing.
$\frac {d}{dq} C = 3$
$R= pq\\ \frac {dR}{dq} = \frac {dp}{dq} q + p = 3\\ \frac {dq}{dp} = -300\\ \frac {dp}{dq} = \frac {1}{-300}\\ \frac {dR}{dq} = \frac {dp}{dq} q + p = 3\\ \frac {1}{-300}q + p = 3$
Everything up to the last line is preliminary and presented here just for background.
Now substitute $q = 4000 - 300 (p-5)$
$\frac {1}{-300}(4000 -300(p-5)) + p = 3\\ 2p + -\frac {5500}{300} = 3\\ p = \frac {55}{6} + \frac {3}{2} = \frac {32}{3}$