Question

During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for \(10 each and his sales averaged 20 per day. When he increased the price by \)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 Terry $6, what should the selling price be to maximize his profit?

Short answer

(a) The demand function when the sales decreases from 20 per day to18 per day for a $1 increase in selling price is \(20 - 0.5x\).

(b) The selling should be set at $13 to maximize the profit

Step-by-step solution

Given data

Average sales per day when the price of necklace was $10 = 20

Average sales per day when the price of necklace was $11 = 18

Maximization condition

The condition for maxima of a function is:

\({\frac{{df\left( x \right)}}{{dx}}_{x = {x_{\max }}}} = 0\;\;\;\;\;.....\left( 1 \right)\)

Demand function

Let the number of sales on a particular day be x. Decrease in sales is:

\(20 - x\)

Price increase per unit change in sales is:

\(\frac{{11 - 10}}{{20 - 18}} = \frac{1}{2}\)

Therefore, demand function is:

\(\begin{aligned}{c}p\left( x \right) &= 10 + \frac{1}{2}\left( {20 - x} \right)\\ &= 10 + 10 - 0.5x\\ &= 20 - 0.5x\end{aligned}\)

The demand function is \(20 - 0.5x\).

Maximum profit

Since the cost to make one necklace is $6, the profit function is:

\(\begin{aligned}{c}P\left( x \right) &= xp\left( x \right) - 6x\\ &= x\left( {20 - 0.5x} \right) - 6x\\ &= 20x - 0.5{x^2} - 6x\\ &= 14x - 0.5{x^2}\end{aligned}\)

The profit is maximum for:

\(\begin{aligned}{c}{\frac{{dP\left( x \right)}}{{dx}}_{x = {x_{\max }}}} &= 0\\14 - {x_{\max }} &= 0\\{x_{\max }} &= 14\end{aligned}\)

The corresponding selling price is:

\(\begin{aligned}{c}p\left( {14} \right) &= 20 - 0.5 \times 14\\ &= 13\end{aligned}\)

The ticket price to maximize profit is $13.