Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 56

Suppose that a man sells a necklace for \$10 each and his sales averaged $\displaystyle 20 \frac{\text{per}}{\text{day}}$. When he increased the price by \$1, he found that the average decreased by two sales per day.
a.) Find the demand function, assuming that it is linear.
b.) If the material for each necklace costs \$6. What should the selling price be to maximize the profit of the man?


a.) Let $P(x)$ be the demand function. If $P(x)$ is linear, it means that the line $y = P(x)$ passes through the point $(20,10)$ and $(18,11)$.
By using Point Slope Form,

$
\begin{equation}
\begin{aligned}
y - 10 &= \frac{11-10}{18-20} (x - 20)\\
\\
y &= - \frac{1}{2}x + 20
\end{aligned}
\end{equation}
$


Therefore, the deman function $\displaystyle P(x) = - \frac{1}{2}x + 20$.


B.) If the demand is $P(x)$, then the revenue $R(x) = x P(x)$
$\displaystyle R(x) = x\left( -\frac{1}{2}x + 20 \right) = -\frac{1}{2}x^2 + 20x$
Thus, we have...

$
\begin{equation}
\begin{aligned}
\text{Profit } &= \text{Revenue } - \text{ cost}\\
\\
P(x) &= R(x) - C(x)\\
\\
P(x) &= -\frac{1}{2}x^2 + 10x - 6x\\
\\
P(x) &= -\frac{1}{2} x^2 + 14x
\end{aligned}
\end{equation}
$


If we take the derivative of $P'(x)$, then...
$P'(x) = -x +14$

when $P'(x) = 0$,
$ x = 14$

Therefore, the selling price for maximum profit is

$
\begin{equation}
\begin{aligned}
P(x) &= - \frac{1}{2} x + 120\\
\\
P(14) &= -\frac{1}{2} (14) + 20\\
\\
P(14) &= \$13
\end{aligned}
\end{equation}
$