Question

During the summer months Terry makes and sells necklaces on the beach. Last summer she sold the necklaces for \(10 each and her sales averaged 20 per day. When she increased the price by \)1, she found that the average decreased by two sales per day.

1. Find the demand function, assuming that it is linear.
2. If the material for each necklace costs $6, what selling price should Terry set to maximize her profit?

Short answer

1. The demand function is\(y = p\left( x \right) = - \frac{x}{2} + 20\)and the revenue is given by\(R\left( x \right) = xp\left( x \right) = - \frac{1}{2}{x^2} + 20x\).
2. The selling price is set to maximize profit is \(\$ 13\).

Step-by-step solution

(a) Step 1: Find the demand function

Consider\(p\left( x \right)\)as the demand function. It is given that the demand function\(p\left( x \right)\)is linear and\(y = p\left( x \right)\)passes through the point\(\left( {20,10} \right)\)and\(\left( {18,11} \right)\).

Obtain the slope as shown below:

\(\begin{aligned}{c}m = \frac{{11 - 10}}{{18 - 20}}\\ = - \frac{1}{2}\end{aligned}\)

Obtain the equation of the line as shown below:

\(\begin{aligned}{c}y - 10 = - \frac{1}{2}\left( {x - 20} \right)\\y = - \frac{x}{2} + \frac{{20}}{2} + 10\\y = - \frac{x}{2} + 10 + 10\\y = - \frac{x}{2} + 20\end{aligned}\)

Thus, the demand function is \(y = p\left( x \right) = - \frac{x}{2} + 20\) and the revenue is given by \(R\left( x \right) = xp\left( x \right) = - \frac{1}{2}{x^2} + 20x\).

(b) Step 2: What selling price should terry set to maximize profit

It is given that the cost is\(C\left( x \right) = 6x\).

The profit is shown below:

\(\begin{aligned}{c}P\left( x \right) = R\left( x \right) - C\left( x \right)\\ = - \frac{1}{2}{x^2} + 20x - 6x\\ = - \frac{1}{2}{x^2} + 14x\end{aligned}\)

Obtain the derivative of\(P\left( x \right)\)as shown below:

\(\begin{aligned}{c}P'\left( x \right) = \frac{d}{{dx}}\left( { - \frac{1}{2}{x^2} + 14x} \right)\\ = - \frac{{2x}}{2} + 14\\ = - x + 14\end{aligned}\)

When\(P'\left( x \right) = 0\)then,

\(\begin{aligned}{c} - x + 14 = 0\\ - x = - 14\\x = 14\end{aligned}\)

The second derivative of the function is shown below:

\(\begin{aligned}{c}P''\left( x \right) = \frac{d}{{dx}}\left( { - x + 14} \right)\\ = - 1\end{aligned}\)

As\(R''\left( x \right) < 0\), the selling price to maximize profit is shown below:

\(\begin{aligned}{c}p\left( {14} \right) = - \frac{1}{2}\left( {14} \right) + 20\\ = - 7 + 20\\ = \$ 13\end{aligned}\)

Thus, the selling price is set to maximize profit is \(\$ 13\).