College Algebra, Chapter 4, 4.1, Section 4.1, Problem 78

A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $\$6$, and the society sells an average of 20 per week at a price of $\$10$ each. The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it loses 2 sales per week.

a.) Find a function that models weekly profit in terms of price per feeder.

b.) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?

a.) If we let $x$ be the price of the feeder and $y$ be the quantity of the feeders sold at price $x$, then by using slope intercept form, then we can now solve the function for the weekly profit. We have two points $(10, 20)$ and $(11, 18)$. The second point is based on the statement that for every dollar increase, the sales loses 2 per week.


$
\begin{equation}
\begin{aligned}

y =& mx + b
\\
\\
y =& \left( \frac{18 - 20}{11 - 10} \right) x + b
\\
\\
y =& -2x + b
\\
\\
& @ (10, 20)
\\
\\
20 =& -2(10) + b
\\
\\
b =& 40


\end{aligned}
\end{equation}
$


Thus,

$y = -2x + 40$

Next, we know that if each feeder cost $\$6$, then the profit from each bird feeder is $x - 6$. Thus, the weekly profit is equal to the profit per bird feeder times the quantities of bird feeders that are sold weekly. So,


$
\begin{equation}
\begin{aligned}

P(x) =& (x - 6)(-2x + 40)
\\
\\
P(x) =& -2x^2 + 52x - 240

\end{aligned}
\end{equation}
$


b.) To maximize the profit, the value of $x$ should be

$\displaystyle x = \frac{-b}{2a} = \frac{-52}{2(-2)} = \$ 13$

Therefore, the maximum weekly profit is


$
\begin{equation}
\begin{aligned}

P(13) =& -2(13)^2 + 52(13) - 240
\\
\\
P(13) =& \$ 98

\end{aligned}
\end{equation}
$