Calculus problem that I've been trying to get my head around.
Problem: A company can sell 20 products if it charges $40 per product.
For each dollar decrease or increase in the price, the company can sell one more or one less product, respectively. The total cost of producing q products is C(q) = 32q + 100. What is the maximum profit that the company can achieve from manufacturing and selling this product?
I tried to find the profit function by subtracting cost from revenue, then took the derivative of said function, but got lost along the process. I know the answer is supposed to be $96.
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$\begingroup$A company can sell 20 products if it charges $40 per product. For each dollar decrease or increase in the price, the company can sell one more or one less product, respectively
That gives the quantity $q$ for the price $p$$$q = 20 + 40 - p$$$$q = 60 - p$$
The total cost of producing q products is $C(q) = 32q + 100$
Thus, the profit $P$ is the quantity $q$ by the price $p$ minus production cost$$P = qp - (32q + 100)$$And, by plugging the value of $q$ above$$ P = p(60 - p) - 32(60 - p) $$$$ P = -p^2 + 92p - 2020 $$
The leading coefficient is negative, meaning the quadratic function is a downward parabola. Thus, finding the price of the curve at the top will give you the maximum profit. That is, for which $p$ is the derivative zero?$$ P' = -2p + 92 $$$$ -2p + 92 = 0 $$$$ p = 46 $$
We know that for a price of $46, we get the maximum profit. Plugging that price into the profit equation gives the maximum profit
$$ P = -p^2 + 92p - 2020 $$$$ maxP = -46^2 + 92\cdot46 - 2020 $$$$ maxP = 96 $$
(image from Wolfram Alpha)
Let $p$ be the price in dollars and $n(p)$ be the number of items sold at price $p$. We know $n(40) = 20$ and $$ n(p) = 20 - (p-40) = 60 - p \text{.} $$(This says that for every dollar we raise the price over $40$, the number of units sold decreases by $1$ and for every dollar we lower the price under $40$, the number of units sold increases by $1$.)
Let $r(p)$ be the revenue at a given price. We have \begin{align*} r(p) &= n(p) \cdot p \\ &= (60 - p) \cdot p \\ &= 60p - p^2 \text{.} \end{align*}We know the cost is $c(n) = 32n+100$. Inserting $n(p)$ we find that the cost of producing the $n(p)$ items sellable at price $p$, we have \begin{align*} c(p) &= 32(60-p) + 100 \\ &= 1920 - 32p \text{.} \end{align*}Let $P(p)$ be the profit at price $p$ if we produce exactly as many items as will be sold at that price. \begin{align*} P(p) &= r(p) - c(p) \\ &= (60p-p^2) - (1920 - 32p) \\ &= -1920 + 92p - p^2 \text{.} \end{align*}Then $$ P'(p) = 92 - 2p $$and this is zero when $p = 46$ at which price, $P(46) = 196$. Checking that this is a maximum,
- using the second derivative: $P''(46) = -2$, so we have a local maximum, or
- using the first derivative: $P'(45) = 2$ and $P'(47) = -2$, so we have a local maximum.
Either way, we should compare with the natural endpoints to make sure there isn't a better optimum at a place where the derivative is not zero. (For instance the maximum and minimum of a non-horizontal line are achieved a the endpoints of the feasible interval.) In this case, we cannot make fewer than $0$ items, corresponding to a price of $60$. $P(60) = 0$, unsurprisingly. Also, we cannot (sanely) price below zero and $P(0) = -1920$, which is far below the maximum found above.
The maximal profit is obtained when the price is \$46, at which price the profit is \$196.
$\endgroup$ $\begingroup$First thing to notice is that you cannot decrease the number of products by more than $n=19$ and you cannot increase it by more that $n=39$. The profit function would be:$$P(n)=(20+n)(40-n)$$Logically, to maximise the net profit is to maximise $NP(n)=P(n)-C(n)$.$$\begin{align*}NP(n)&=P(n)-C(n)\\&=(20+n)(40-n)-32(20+n)-100\\&=-n^2-12n+60\end{align*}$$This is the equation of a concave down parabola, so it has a maximum.$$\begin{align*} \frac{\mathrm dNP(n)}{\mathrm{d}n}=0&\Leftrightarrow -2n-12=0\\&\Leftrightarrow n=-6\text{ verifies } -19\leq n \leq 39 \end{align*}$$You need to decrease the number of products by $6$ and the net profit would be $NP(6)=96\,\$$.
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